Ta có \(y'=\left(x^3-\displaystyle\frac{1}{2}x^2+3x-5\right)' = 3x^2-x+3.\)

Ta có \(y'=(x+1)'(x-2)+(x+1)(x-2)'=x-2+x+1=2x-1\).

Ta có \(y'=15x^4-2x^3+9x^2+2\).

Ta có \(\left(\displaystyle\frac{1}{3}x^3+x^2-3x \right)'=x^2+2x-3\), suy ra \(a=1\), \(b=2\), \(c=-3\).

Vậy \(S=a+2b+3c=1+2\cdot 2+3\cdot (-3)=-4\).

Ta có

\begin{eqnarray*}f'(x)&=&2\left(x^3-2x^2\right) (x^3-2x^2)'\\&=&2\left(3x^2-4x\right)\left(x^3-2x^2\right)\\&=&2\left(3x^5-4x^4-6x^4+8x^3\right)\\&=&6x^5-20x^4+16x^3.\end{eqnarray*}

Ta có: \(y=\displaystyle\frac{1}{3}x^3+1\Rightarrow y'=\displaystyle\frac{1}{3}\cdot 3x^2+0=x^2\).

Ta có \(y'=(2x^5-x^2+3)'=10x^4-2x\).

Ta có \(y = 2x(x^6 + 1) = 2x^7 + 2x\) nên \(y' = 14x^6 + 2\).

Ta có \(y=(5 x-1)^{2}\Rightarrow y'=10(5x-1)=50x-10\).

Ta có \(y'=3x^2+4x\).

Hàm số \(y=x^7-3x^5+4x^3+3\) có đạo hàm \(y'=7x^6-15x^4+12x^2\).

\(y'=-6x^5+6x^2-2\)

Hàm số \(y=\displaystyle\frac{x^4}{2} -\displaystyle\frac{x^3}{3} + x+2\) có đạo hàm là

\[y'=\left(\displaystyle\frac{x^4}{2} -\displaystyle\frac{x^3}{3} + x+2\right)'=4\cdot \displaystyle\frac{x^3}{2}-3\cdot \displaystyle\frac{x^2}{3}+1=2x^3-x^2+1.\]

Ta có \(y'=3x^2-2x+5\).

Ta có \(y'=6x^2.\)

Ta có \(y'=(3x^2+4x)'=6x+4\).

Đạo hàm \(y'=(-x^3)'=-3x^2\).

Ta có \(y' = 4x^3 - 6x + 2\).

Ta có \(y'=4x^3-6x+3\).

Ta có \(y'=-\displaystyle\frac{1}{3}+2x-x^3\).

Ta có \(f'(x) =3x^2 +4x-7\).

\(y'=6x^2 \Rightarrow y'(-1)=6.\)

Ta có \(f'(x)=3x^2-6x\).

Suy ra \(f'(3)=3\cdot 3^2-6\cdot 3=9\).

Ta có \(y'=-3x^2+3\) nên \(y'(1)=-3\cdot 1^2+3=0\).

Ta có \(f'(x)=4x^3+4x\). Suy ra \(f'(-1)=4\cdot (-1)^3+4\cdot (-1)=-8\).

Ta có \(f'(x)=4x^3-6x\Rightarrow f'(2)=20.\)

Ta có \(f'(x)=3x^2-6x\), do đó \(f'(2)=0\).

Ta có \(y=f(x)=x^2+x+1 \Rightarrow y'=f'(x)=2x+1 \Rightarrow f'(1)=2\cdot 1 +1=3.\)

Ta có \(f'(x)=-9x^2\Rightarrow f'(-1)=-9\).

Tập xác định: \(\mathscr{D}=\mathbb{R}\setminus\{0\}\).

Do \(y'(x)=14x^6+\displaystyle\frac{3}{x^2}+2\) nên \(y'(-1)=19\).

Tập xác định: \(\mathscr{D}=\mathbb{R}\).

Do \(y'(x)=3x^2\) nên \(y'(2)=12\).

\(f'(x)=2ax+2b\).

\(f'(1)=6 \Leftrightarrow 2a+2b=6 \Leftrightarrow a+b=3\).

Ta có \(f'(x)=3x^2-6x \Rightarrow f'(1)=3-6=-3\).

Ta có \(y'=f'(x)=2x+2\Rightarrow f'(1)=2\cdot 1+2=4\).

Ta có: \(f'(x)=-4x^3+12x^2-6x+2\).

Suy ra \(f'(-1)=-4{{(-1)}^3}+12{{(-1)}^2}-6(-1)+2=24\).

Ta có \(f'(x)=x^2+4mx+3\).

Suy ra \(f'(1)=1^2+4m\cdot 1+3=4+4m\).

Ta có \(y'=6x^2\Rightarrow y'(2)=6\cdot 2^2=24\).

Ta có \(\left(x^5\right)^\prime =5 x^4\). Từ đó, \(y^\prime \left(2\right)=5\cdot 2^4=80\) và \(y^\prime \left(-\displaystyle\frac{1}{2}\right)=5\cdot \left(-\displaystyle\frac{1}{2}\right) ^4=\displaystyle\frac{5}{16}\).

Ta có \(\left(x^{10}\right)^\prime =10 x^9\). Từ đó, \(y^\prime \left(-1\right)=10\cdot \left(-1\right)^9=-10\) và \(y^\prime \left(\sqrt[3]{2}\right)=10\cdot \left(\sqrt[3]{2}\right) ^9=80\).

Ta có \(f'(x)=x^2-4\sqrt{2}x+8\).

\(f'(x)=0\Leftrightarrow x=2\sqrt{2}\).

Ta có \(f'(x)=\left(-x^3+2x^2-x+5\right)'=-3x^2+4x-1\).

Suy ra \(f'(x)=0\Leftrightarrow -3x^2+4x-1=0\Leftrightarrow \hoac{&x=1;\,x=\displaystyle\frac{1}{3}.}\)

Vậy \(S=\left\{1;\displaystyle\frac 13\right\}\).

Ta có \(f'(x)=\left(-x^4+3x^2+2\right)'=-4x^3+6x\).

Khi đó \(f'(x)=0\Leftrightarrow -4x^3+6x=0 \Leftrightarrow \hoac{&x=0;\,x=\displaystyle\frac{\sqrt{6}}{2};\,x=-\displaystyle\frac{\sqrt{6}}{2}.}\)

Vậy phương trình \(f'(x)=0\) có \(3\) nghiệm.

Ta có \(y'<0\Leftrightarrow 3x^2-2x-1<0\Leftrightarrow -\displaystyle\frac{1}{3}

Vậy \(x\in\left(-\displaystyle\frac{1}{3};1\right)\).

Ta có \(f'(x) = x^2 - 3x^2 - 4\) nên \(f'(x) = 0 \Leftrightarrow x^2 - 3x^2 - 4 = 0 \Leftrightarrow \hoac{&x = -1;\,x = 4.}\)

Ta có \(y'=3x^2-6x-9\).

Suy ra \(y'>0\Leftrightarrow \hoac{&x<-1;\,x>3.}\)

Vậy tập nghiệm của bất phương trình \(y'>0\) là \((-\infty ;-1) \cup(3 ;+\infty)\).

\(f'(x) = 0 \Leftrightarrow x^2 - 3x -4 = 0 \Leftrightarrow x = -1 \vee x = 4\).

Ta có \(f'(x)=-x^{2}+7x-5\). Theo Viet thì \(x_{1}+x_{2}=-7\).

Ta có: \(f'(x)=x^2-4x-5\).

Do đó \(f'(x)=0 \Leftrightarrow x^2-4x-5=0 \Leftrightarrow \left[\begin{aligned}&x=-1 \\ &x=5\end{aligned}\right.\).

Ta có \(f'(x)= x^2-3x+2\). Do đó \(f'(x)=0 \Leftrightarrow x^2-3x+2=0 \Leftrightarrow \hoac{&x=1\\ &x=2} \Rightarrow S=\{1; 2\}\).

Ta có \(f'(x)=3x^2-6x\).

Khi đó \(f'(x)<0\Leftrightarrow 3x^2-6x<0\Leftrightarrow x\in (0;2)\).

Ta có \(f'(x)=3x^2-6x\) \(\Rightarrow f'(x)<0 \Leftrightarrow 3x^2-6x<0 \Leftrightarrow \hoac{&x<0;\,x>2}\).

Với \(\forall x \in \mathbb{R}\), ta có \(y'=3x^2-6x\).

Khi đó \(y'=0 \Leftrightarrow 3x^2-6x=0 \Leftrightarrow \hoac{& x=0 ;\, x=2.}\)

Vậy tổng các nghiệm của phương trình \(y'=0\) bằng \(2\).

Ta có \(f'(x)=3x^2-6x\). Khi đó \(f'(x)\leq 0\Leftrightarrow 3x^2-6x\leq 0\Leftrightarrow x\in[0;2].\)

Vậy tập nghiệm của bất phương trình \(f'(x)\leq 0\) là \(S=[0;2]\).

Ta có \(f'(x)=3x^2-12x\). Do đó

\[f'(x)<0\Leftrightarrow 3x^2-12x<0\Leftrightarrow x\in (0;4).\]

Ta có \(\left(\displaystyle\frac{1}{u}\right)'=-\displaystyle\frac{1}{u^2}\cdot u'\).

Vậy \(y'=-\displaystyle\frac{1}{x^4}\cdot (x^2)'=-\displaystyle\frac{1}{x^4}\cdot (2x)=-\displaystyle\frac{2}{x^3}\).

Ta có \(f'(x)=5x^4+\displaystyle\frac{1}{x^2}\). Suy ra \(f'(1)=5+1=6\).

Ta có \(y'=\left(x-\displaystyle\frac{4}{x}\right)'=1+\displaystyle\frac{4}{x^2}=\displaystyle\frac{x^2+4}{x^2}\).

Ta có \(y'=\left(x+\displaystyle\frac{9}{x}\right)'=1-\displaystyle\frac{9}{x^{2}}\).

Ta có \(y'=\displaystyle\frac{4}{2\sqrt{x}}+\displaystyle\frac{5}{x^2}=\displaystyle\frac{2}{\sqrt{x}}+\displaystyle\frac{5}{x^2}\).

Ta có \(f'(x)=-\displaystyle\frac{1}{x^2}+\displaystyle\frac{1}{2}\cdot \displaystyle\frac{-2x}{x^4}+\displaystyle\frac{1}{3}\cdot\displaystyle\frac{-3x^2}{x^6}=\displaystyle\frac{1}{-x^2}+\displaystyle\frac{1}{-x^3}+\displaystyle\frac{1}{-x^4}\).

Vậy \(a+b+c=-1+(-1)+(-1)=-3\).

Ta có \(y'=2x-\displaystyle\frac{3}{2\sqrt{x}}-\displaystyle\frac{1}{x^2}\).

Ta có: \(y'=6x+\displaystyle\frac{1}{\sqrt{x}}\).

Ta có \(y'=-\displaystyle\frac{\left(x\sqrt{x}\right)'}{\left(x\sqrt{x}\right)^2}=-\displaystyle\frac{\sqrt{x}+x\cdot\displaystyle\frac{1}{2\sqrt{x}}}{x^3}=-\displaystyle\frac{\tfrac{3}{2}\sqrt{x}}{x^3}=-\displaystyle\frac{3}{2x^2\sqrt{x}}\).

\(f'(x)=(x^2+\sqrt{x})=(x^2)'+(\sqrt{x})'=2x+\displaystyle\frac{1}{2\sqrt{x}}\).

Ta có

\(y=x^{\frac{2}{3}}\)

\(y'=\displaystyle\frac{2}{3}\cdot x^{-\frac{1}{3}}=\displaystyle\frac{2}{3}\cdot\displaystyle\frac{1}{\sqrt[3]{x}}\).

Hàm số \(f(x)=x^3-\displaystyle\frac{1}{x}+3x\) có đạo hàm là \(f'(x)=3 x^2+\displaystyle\frac{1}{x^2}+3\), \(\forall x\not=0\).

Ta có \(y'=(3x^2+\sqrt{x})'=6x+\displaystyle\frac{1}{2\sqrt{x}}\).

Ta có \(f'(x)=-4x^3+\displaystyle\frac{2}{2\sqrt{x}}=-4x^3+\displaystyle\frac{1}{\sqrt{x}}\).

Ta có \(y' = \left(x^{2}\right)'-\left(3 \sqrt{x}\right)'+\left(\displaystyle\frac{1}{x}\right)' = 2x -\displaystyle\frac{3}{2\sqrt{x}} - \displaystyle\frac{1}{x^2}\).

Ta có \(y'=-14x^6+\displaystyle\frac{1}{2\sqrt{x}}\).

Ta có \(y'=\left(5-4\sqrt{x}\right)'=-4\cdot\displaystyle\frac{1}{2\sqrt{x}}=-\displaystyle\frac{2}{\sqrt{x}}\).

Ta có \(y'=4x^3-\displaystyle\frac{1}{\sqrt{x}}\).

\(y'=5x^4+\displaystyle\frac{3}{2\sqrt{x}} \Rightarrow y'(1)=5 \cdot 1^4+\displaystyle\frac{3}{2\sqrt{1}}=\displaystyle\frac{13}{2}\).

Ta có \(y'=2\cdot \displaystyle\frac{1}{2\sqrt{x}}+1=\displaystyle\frac{1}{\sqrt{x}}+1\).

Vậy \(y'(1)=\displaystyle\frac{1}{\sqrt{1}}+1=2\).

Ta có \(f'(x) = 2x+ \displaystyle\frac{4}{2 \sqrt{x}} = 2x +\displaystyle\frac{2}{\sqrt{x}}\) với mọi \(x >0\). Suy ra \(f'(4) = 2 \cdot 4 + \displaystyle\frac{2}{ \sqrt{4}}=9\).

Ta có \(y'=\displaystyle\frac{4}{2\sqrt{x}}=\displaystyle\frac{2}{\sqrt{x}}\Rightarrow y'(4)=1\).

Ta có \(f(x)=\displaystyle\frac{x\sqrt{x}+1}{x}=\sqrt{x}+\displaystyle\frac{1}{x}\) với \(x>0\). Do đó

\[f'(x)=\left(\sqrt{x}+\displaystyle\frac{1}{x}\right)' = \left(\sqrt{x}\right)'+\left(\displaystyle\frac{1}{x}\right)' = \displaystyle\frac{1}{2\sqrt{x}}-\displaystyle\frac{1}{x^2}.\]

Ta có \(y'=\displaystyle\frac{(x-1)'(2x-1)-(2x-1)'(x-1)}{(2x-1)^2}=\displaystyle\frac{1}{(2x-1)^2}\).

Ta có \(y'=\displaystyle\frac{(2x-1)'(x+2)-(x+2)'(2x-1)}{(x+2)^2}=\displaystyle\frac{2x+4-2x+1}{(x+2)^2}=\displaystyle\frac{5}{(x+2)^2}.\)

Ta có \(y'=\left(\displaystyle\frac{2x-1}{x+3}\right)' = \displaystyle\frac{7}{(x+3)^2}.\)

Áp dụng công thức đạo hàm nhanh \({\left(\displaystyle\frac{ax+b}{cx+d}\right)}'=\displaystyle\frac{ad-bc}{(cx+d)^2} \Rightarrow {\left(\displaystyle\frac{-2x+1}{-x+1}\right)}'=-\displaystyle\frac{1}{(1-x)^2}\).

\(y'=\displaystyle\frac{-4(2x-5)-2(3-4x)}{(2x-5)^2}=\displaystyle\frac{-8x+20-6+8x}{(2x-5)^2}=\displaystyle\frac{14}{(2x-5)^2}\).

\(y=\displaystyle\frac{x(1+2x)}{1-x}=\displaystyle\frac{x+2x^2}{1-x}\)

\(y'=\displaystyle\frac{(x+2x^2)'(1-x)-(x+2x^2)(1-x)'}{(1-x)^2}=\displaystyle\frac{(1+4x)(1-x)-(x+2x^2)(-1)}{(1-x)^2}=\displaystyle\frac{-2x^2+4x+1}{(1-x)^2}\).

\(y' =\displaystyle\frac{(3-x)'(2x+1)-(3-x)(2x+1)'}{(2x+1)^2}=\displaystyle\frac{-2x-1-6+2x}{(2x+1)^2}=\displaystyle\frac{-7}{(2x+1)^2}\).

Ta có \(y'=\displaystyle\frac{2(3x-2)-3(2x-3)}{(3x-2)^2}=\displaystyle\frac{5}{(3x-2)^2}\).

Tập xác định: \(\mathscr{D}=\mathbb{R}\setminus\{1\}\).

Ta có \(y'=\displaystyle\frac{2(x-1)-(2x+1)}{(x-1)^2}=\displaystyle\frac{-3}{(x-1)^2}\).

Ta có \(y'=\displaystyle\frac{7}{\left(x+2\right)^2}=\displaystyle\frac{a}{(x+2)^2}\Rightarrow a=7\).

Vậy \(P=a^2-3=46\).

\(y'=\displaystyle\frac{(2x+1)'(x+2)-(2x+1)(x+2)'}{(x+2)^2}=\displaystyle\frac{2x+4-2x-1}{(x+2)^2}=\displaystyle\frac{3}{(x+2)^2}\).

\(\Rightarrow a=3\).

Ta có \(y' = \left(\displaystyle\frac{2x - 1}{1 - x} \right)' = \left(\displaystyle\frac{2x - 1}{-x + 1} \right)' = \displaystyle\frac{2\cdot 1 - (-1)\cdot (-1)}{(-x + 1)^2} = \displaystyle\frac{1}{(x - 1)^2}\).

Tập xác định của hàm số là \(\mathbb{R}\setminus\left\{\displaystyle\frac{1}{4}\right\}\).

Trên mỗi khoảng \(\left(-\infty;\displaystyle\frac{1}{4}\right)\); \(\left(\displaystyle\frac{1}{4};+\infty\right)\), ta có

\(y'=\left(\displaystyle\frac{1}{1-4x}\right)'=\displaystyle\frac{4}{(4x-1)^2}=\displaystyle\frac{4}{(1-4x)^2}.\)

Ta có \(y'=\displaystyle\frac{(2x+1)(x+1)-\left(x^2+x-2\right)}{(x+1)^2}=\displaystyle\frac{x^2+2x+3}{(x+1)^2}\).

Do đó \(a=2\), \(b=3\) \(\Rightarrow P=a \cdot b=6\).

Ta có \(y'=\displaystyle\frac{3(2x+1)-2(3x+1)}{(2x+1)^2}=\displaystyle\frac{1}{(2x+1)^2}\).

Ta có \(y'=\displaystyle\frac{(2x-3)'\cdot(x+4)-(x+4)'\cdot(2x-3)}{(x+4)^2}=\displaystyle\frac{2(x+4)-(2x-3)}{(x+4)^2}=\displaystyle\frac{11}{(x+4)^{2}}\).

\(y'=-\displaystyle\frac{3}{(x-1)^2}\).

Tập xác định \(\mathscr{D}=\mathbb{R}\setminus \left\lbrace 1 \right\rbrace\).

\(\begin{aligned}y'&=\displaystyle\frac{(2x+1)'(1-x)-(1-x)'(2x+1)}{(1-x)^2}\\&=\displaystyle\frac{2(1-x)-(-1)(2x+1)}{(1-x)^2}\\&=\displaystyle\frac{2-2x+2x+1}{(1-x)^2}\\&=\displaystyle\frac{3}{(1-x)^2}.\end{aligned}\)

Vậy \(y'=\displaystyle\frac{3}{(1-x)^2}.\)

Ta có \(f'(x)= \displaystyle\frac{(x-1)'(x+2)-(x-1)(x+2)'}{(x+2)^2}=\displaystyle\frac{1\cdot (x+2)-(x-1)\cdot 1}{(x+2)^2} =\displaystyle\frac{3}{(x+2)^2}\).

Suy ra \(f'(2)= \displaystyle\frac{3}{(2+2)^2}= \displaystyle\frac{3}{16}\).

Ta có \(f'(x)=\displaystyle\frac{-2}{(x-1)^2}\Rightarrow f'(-1)=\displaystyle\frac{-2}{(-1-1)^2}=\displaystyle\frac{-2}{4}=-\displaystyle\frac{1}{2}\).

Ta có \(f'(x)=\displaystyle\frac{-7}{(2x-1)^{2}}\Rightarrow f'(1)=-7\).

Ta có \(y'=\displaystyle\frac{-5}{(1+x)^2}\) suy ra \(y'(0) = -5\).

Ta có

\(y'=\displaystyle\frac{(2x-2)(x-2)-(x^2-2x-1)}{(x-2)^2}=\displaystyle\frac{x^2-4x+5}{(x-2)^2}.\)

Ta có \(y'=\displaystyle\frac{(2x-3)(x^2+x-2)-(x^2-3x+4)(2x+1)}{\left(x^2+x-12\right)^{2}}=\displaystyle\frac{4x^2-12x+2}{(x^2+x-12)^2}.\)

\(f'(x)=\displaystyle\frac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}=\displaystyle\frac{4x}{(x^2+1)^2}\).

\(f'(x)=0\Leftrightarrow x=0\).

Ta có \(y' = \displaystyle\frac{(x^2)'\cdot (x-1) - (x-1)'\cdot x^2}{(x-1)^2} = \displaystyle\frac{ 2x(x-1) - x^2 }{(x-1)^2} = \displaystyle\frac{x^2 - 2x}{(x-1)^2}\).

Do đó \(y'=0\Leftrightarrow \begin{cases}x^2-2x=0\\ x\ne1\end{cases}\Leftrightarrow x\in\{0;2\}\).

Ta có \(y'=\displaystyle\frac{-x^2+4x-1}{(2-x)^2}\), suy ra \(a=-1\), \(b=4\) và \(c=-1\).

Vậy \(S=a+b+c=2\).

Ta có \(y'=\displaystyle\frac{(2x+1)(x+1)-(x^2+x-2)}{(x+1)^2}=\displaystyle\frac{x^2+2x+3}{(x+1)^2}\).

Khi đó \(a=2\), \(b=3\Rightarrow P=2a+b=7.\)

Ta có

\begin{align*}y' &= \displaystyle\frac{(x^2+x+1)' \cdot (x+1) - (x^2+x+1) \cdot (x+1)'}{(x+1)^2} \\&= \displaystyle\frac{(2x+1)(x+1) - (x^2+x+1)}{(x+1)^2} \\& = \displaystyle\frac{x^2+2x}{(x+1)^2}. \end{align*}

Ta có \begin{eqnarray*}y'&=&\displaystyle\frac{(x+1)(x^2+2x+2)'-(x^2+2x+2)(x+1)'}{(x+1)^2}\\&=&\displaystyle\frac{(x+1)(2x+2)-(x^2+2x+2)\cdot1}{(x+1)^2}\\&=&\displaystyle\frac{2x^2+4x+2-x^2-2x-2}{(x+1)^2}\\&=&\displaystyle\frac{x^2+2x}{(x+1)^2}.\end{eqnarray*}

Ta có \(y'= \displaystyle\frac{(2x+1)(x+1)-\left(x^2+x+1\right)}{ (x+1)^2}= \displaystyle\frac{x^2+2x}{(x+1)^2}\).

Ta có \(y'=\displaystyle\frac{2x^2-8x+1}{(x-2)^2}\).

\(y'=0\Leftrightarrow 2x^2-8x+1=0\).\hfill\((*)\)

Tổng các nghiệm của phương trình \((*)\) là \(S=x_1+x_2=-\displaystyle\frac{b}{a}=\displaystyle\frac{8}{2}=4\).

Ta có \(y' =\displaystyle\frac{(4x+2)(x^2+x+3)-(2x+1)(2x^2+2x+3)}{(x^2+x+3)^2}=\displaystyle\frac{6x+3}{(x^2+x+3)^2}\).

Ta có \(y=x-\displaystyle\frac{3}{x+2}\Rightarrow y'=1+\displaystyle\frac{3}{(x+2)^2}\).

Ta có \(y=\displaystyle\frac{x(1-3x)}{x+1}=\displaystyle\frac{x-3x^2}{x+1}\).

Suy ra \(y'=\displaystyle\frac{(x-3x^2)'(x+1)-(x-3x^2)(x+1)'}{(x+1)^2}=\displaystyle\frac{(1-6x)(x+1)-(x-3x^2)}{(x+1)^2}=\displaystyle\frac{-3x^2-6x+1}{(x+1)^2}.\)

Ta có

\begin{eqnarray*}f'(x)&=&\displaystyle\frac{(1-3x+x^2)'(x-1)-(1-3x+x^2)(x-1)'}{(x-1)^2}\\ &=&\displaystyle\frac{(-3+2x)(x-1)-(1-3x+x^2)}{(x-1)^2}\\&=&\displaystyle\frac{x^2-2x+2}{(x-1)^2}.\end{eqnarray*}

Bất phương trình \(f'(x)>0\Leftrightarrow \displaystyle\frac{x^2-2x+2}{(x-1)^2}>0\Leftrightarrow \begin{cases} x^2-2x+2>0 \\ x\ne 1\end{cases}\Leftrightarrow x\in \mathbb{R}\setminus \{1\}\).

Ta có \(f'(x)=\displaystyle\frac{(x^3)'(x-1)-x^3(x-1)'}{(x-1)^2}=\displaystyle\frac{3x^2(x-1)-x^3}{(x-1)^2}=\displaystyle\frac{2x^3-3x^2}{(x-1)^2}.\)

Phương trình \(f'(x)=0\Leftrightarrow \displaystyle\frac{2x^3-3x^2}{(x-1)^2}=0\Leftrightarrow 2x^3-3x^2=0\Leftrightarrow x=0 ;\, x=\displaystyle\frac{3}{2}.\)

Ta có \(y'=\displaystyle\frac{-(x^2-2x+5)'}{(x^2-2x+5)^2}=\displaystyle\frac{-2x+2}{(x^2-2x+5)^2}. \)

Ta có

\begin{eqnarray*}y'&=&\displaystyle\frac{(2x+5)'(x^2+3x+3)-(2x+5)(x^2+3x+3)'}{(x^2+3x+3)^2}\\&=&\displaystyle\frac{2(x^2+3x+3)-(2x+5)(2x+3)}{(x^2+3x+3)^2}\\&=&\displaystyle\frac{-2x^2-10x-9}{(x^2+3x+3)^2}.\end{eqnarray*}

Ta có

\begin{eqnarray*}y'&=&\displaystyle\frac{(-2x^2+x-7)'(x^2+3)-(x^2+3)'(-2x^2+x-7)}{(x^2+3)^2}\\&=&\displaystyle\frac{(-4x+1)(x^2+3)-2x(-2x^2+x-7)}{(x^2+3)^2}=\displaystyle\frac{-x^2+2x+3}{(x^2+3)^2.}\end{eqnarray*}

Ta có \(f(x)=\displaystyle\frac{x^2-2x+5}{x-1}=x-1+\displaystyle\frac{4}{x-1}\Rightarrow f'(x)=1-\displaystyle\frac{4}{(x-1)^2}\).

Vậy \(f'(2)=1-\displaystyle\frac{4}{(2-1)^2}=-3\).

Ta có \(f'(x)=\displaystyle\frac{(x^2+x)'(x-2)-(x^2+x)(x-2)'}{(x-2)^2}=\displaystyle\frac{(2x+1)(x-2)-(x^2+x)}{(x-2)^2}=\displaystyle\frac{x^2-4x-2}{(x-2)^2}.\)

Suy ra \(f'(1)=-5\).

Ta có \(y'=\displaystyle\frac{\left(x^2+2x\right)'(x-2)-\left(x^2+2x\right)(x-2)'}{(x-2)^2}=\displaystyle\frac{2x^2-2x-4-x^2-2x}{(x-2)^2}=\displaystyle\frac{x^2-4x-4}{(x-2)^2}\).

Vậy \(y'(1)=-7\).

Ta có \(y=f(x)=\displaystyle\frac{3x^4-2x^3+7}{x}=3x^3-2x^2+\displaystyle\frac{7}{x}\) nên \(y'=f'(x)=9x^2-4x-\displaystyle\frac{7}{x^2}\).

Do đó \(f'(-1)=9+4-7=6\).

Ta có

\(y'=\displaystyle\frac{(1-2x^2)'}{2\sqrt{1-2x^2}}=\displaystyle\frac{-4x}{2\sqrt{1-2x^2}}=\displaystyle\frac{-2x}{\sqrt{1-2x^2}}.\)

Ta có \(y'=\displaystyle\frac{2x-12x^2}{2\sqrt{x^2-4x^3}}=\displaystyle\frac{x-6x^2}{\sqrt{x^2-4x^3}}\).

Ta có \(y' = \left(\sqrt{x^2+1} \right)' = \displaystyle\frac{(x^2+1)'}{2\sqrt{x^2+1}} = \displaystyle\frac{2x}{2\sqrt{x^2+1}} = \displaystyle\frac{x}{\sqrt{x^2+1}}\).

\(f'(x)=\displaystyle\frac{(x^2+2x)'}{2\sqrt{x^2+2x}}=\displaystyle\frac{2x+2}{2\sqrt{x^2+2x}}=\displaystyle\frac{x+1}{\sqrt{x^2+2x}}\).

Ta có: \(y' = \left(\sqrt{x^2 + 1} - x \right)' = \displaystyle\frac{x}{\sqrt{x^2 + 1}} - 1\).

Ta có \(y' = \left(\sqrt{x^2-4x}\right)'=\displaystyle\frac{x-2}{\sqrt{x^2-4x}}.\)

Ta có \(y'=\displaystyle\frac{2x-2}{2\sqrt{x^{2}-2x+5}}=\displaystyle\frac{x-1}{\sqrt{x^{2}-2x+5}}\).

\(y'=\displaystyle\frac{(4x^2+1)'}{2\sqrt{4x^2+1}}=\displaystyle\frac{4x}{\sqrt{4x^2+1}}\).

\(y'=\displaystyle\frac{(3x^2-4x+5)'}{2\sqrt{3x^2-4x+5}}=\displaystyle\frac{6x-4}{2\sqrt{3x^2-4x+5}}=\displaystyle\frac{3x-2}{\sqrt{3x^2-4x+5}}\).

Ta có \(y'=-\displaystyle\frac{\left(\sqrt{x^2+1} \right)'}{x^2+1} = -\displaystyle\frac{x}{(x^2+1)\sqrt{x^2+1}}\).

Ta có \(y'=\displaystyle\frac{(4x-x^2)'}{2\sqrt{4x-x^2}}=\displaystyle\frac{4-2x}{2\sqrt{4x-x^2}}=\displaystyle\frac{2-x}{\sqrt{4x-x^2}}\).

Điều kiện \(x^2+x+1\ge 0\) (luôn đúng).

\(y'=\displaystyle\frac{(x^2+x+1)}{2\sqrt{x^2+x+1}}=\displaystyle\frac{2x+1}{2\sqrt{x^2+x+1}}\).

\(y'>0\Leftrightarrow\displaystyle\frac{2x+1}{2\sqrt{x^2+x+1}}>0\Leftrightarrow 2x+1>0\Leftrightarrow x>-\displaystyle\frac{1}{2}\).

Ta có \(y=\sqrt{2 x-2}\) \((x>1)\Rightarrow y' = \displaystyle\frac{1}{2 \sqrt{2x-2}} \cdot (2x-2)' = \displaystyle\frac{2}{2 \sqrt{2x-2}} = \displaystyle\frac{1}{ \sqrt{2x-2}}\).

Ta có \(y'=\left(\sqrt{x^2+1}\right)'=\displaystyle\frac{\left(x^2+1\right)'}{2\sqrt{x^2+1}}=\displaystyle\frac{2x}{2\sqrt{x^2+1}}=\displaystyle\frac{x}{\sqrt{x^2+1}}\).

Ta có \(y'=\displaystyle\frac{(1-3x^2)'}{2\sqrt{1-3x^2}}=\displaystyle\frac{-6x}{2\sqrt{1-3x^2}}=\displaystyle\frac{-3x}{\sqrt{1-3x^2}}\).

Ta có \(f'(x)=\displaystyle\frac{\left(4x^2+8x+5\right)'}{2\sqrt{4x^2+8x+5}}=\displaystyle\frac{8x+8}{2\sqrt{4x^2+8x+5}}=\displaystyle\frac{4x+4}{\sqrt{4x^2+8x+5}}\).

Vậy \(a+b=4+4=8\).

Ta có \(y'=\left(\sqrt{3x+2}\right)'= \displaystyle\frac{(3x+2)'}{2\sqrt{3x+2}}=\displaystyle\frac{3}{2 \sqrt{3x+2}}.\)

Điều kiện xác định \(3-2 x^{2} \ge 0\).

Ta có \(y'=\displaystyle\frac{-4x}{2\sqrt{3-2x^2}}=\displaystyle\frac{-2x}{\sqrt{3-2x^2}}\).

Ta có \[y'=\displaystyle\frac{1}{2\sqrt{2x^2-6x+11}}\cdot \left(2x^2-6x+11\right)' = \displaystyle\frac{1}{2\sqrt{2x^2-6x+11}}\cdot \left(4x-6\right) = \displaystyle\frac{2x-3}{\sqrt{2x^2-6x+11}}.\]

Suy ra \(c=2\) và \(d=-3\).

Ta có \(y'=\displaystyle\frac{\left(1-2x^2\right)'}{2\sqrt{1-2x^2}}=-\displaystyle\frac{4x}{2\sqrt{1-2x^2}}=-\displaystyle\frac{2x}{\sqrt{1-2x^2}}\).

Vậy \(y'=-\displaystyle\frac{2x}{\sqrt{1-2x^2}}\).

Ta có \(y' = \displaystyle\frac{\left(3x^{2}-2x+1\right)'}{2\sqrt{3x^{2}-2x+1}}=\displaystyle\frac{6x-2}{2\sqrt{3x^{2}-2x+1}}=\displaystyle\frac{3x-1}{\sqrt{3x^{2}-2x+1}}\).

Ta có

\(y'=\displaystyle\frac{(x^3-3x^2)'}{2\sqrt{x^3-3x^2}}=\displaystyle\frac{3x^2-6x}{2\sqrt{x^3-3x^2}}\).

Áp dụng công thức ta có \(y'=\displaystyle\frac{(x^2+1)'}{2 \sqrt{x^2+1}}=\displaystyle\frac{2x}{2 \sqrt{x^2+1}}=\displaystyle\frac{x}{\sqrt{x^2+1}}.\)

Ta có \(y'=\displaystyle\frac{-3x}{\sqrt{1-3x^2}}\).

Ta có đạo hàm của hàm số \(f(x)=\sqrt{2-3 x^2}\) là

\(f'(x)=\displaystyle\frac{\left(2-3x \right)^2 }{2\sqrt{2-3 x^2}}=\displaystyle\frac{-3 x}{ \sqrt{2-3 x^2}}\).

Ta có \(f'(x)=\displaystyle\frac{1}{2\sqrt{x+1}}\). Do đó \(f'(3)=\displaystyle\frac{1}{2\sqrt{3+1}} = \displaystyle\frac{1}{2 \cdot 2} = \displaystyle\frac{1}{4}\).

Ta có \(f(1)=\sqrt{8+1}=3\), \(f'(x)=\displaystyle\frac{1}{2\sqrt{8+x}}\Rightarrow f'(1)=\displaystyle\frac{1}{6}\).

Vậy \(f(1)+12f'(1)=3+12\cdot \displaystyle\frac{1}{6}=5\).

Ta có \(f'(x)=\displaystyle\frac{1}{2\sqrt{x+1}}\). Suy ra \(f'(3)=\displaystyle\frac{1}{2\sqrt{3+1}}=\displaystyle\frac{1}{4}\).

Ta có \(y'=\left(\sqrt{10x-x^2}\right)'=\displaystyle\frac{(10x-x^2)'}{2\sqrt{10x-x^2}}=\displaystyle\frac{10-2x}{2\sqrt{10x-x^2}}=\displaystyle\frac{5-x}{\sqrt{10x-x^2}}\Rightarrow y'(2)=\displaystyle\frac{3}{4}\).

Ta có \(f(x)=\sqrt{3+x} \Rightarrow f'(x)=\displaystyle\frac{1}{2\sqrt{3+x}}\).

Nên \(f(1)+4f'(1)=\sqrt{3+1}+\displaystyle\frac{4}{2\sqrt{3+1}}=3\).

Ta có \(f(1)=\sqrt{3+1}=2\) và \(f'(x)=\displaystyle\frac{(3+x)'}{2\sqrt{3+x}}=\displaystyle\frac{1}{2\sqrt{3+x}}\Rightarrow f'(1)=\displaystyle\frac{1}{4}\).

Vậy \(f(1)+4f'(1)=2+4\cdot\displaystyle\frac{1}{4}=3\).

Ta có \(f' ( x )=(\sqrt{x-1})'= \displaystyle\frac{1}{ 2 \sqrt{x-1}}\).

Thế \(x=2\) ta được \(f'(2)=1\).

Ta có \(f'(x)=\displaystyle\frac{1}{2\sqrt{x-1}}.\)

\(f'(2)=\displaystyle\frac{1}{2\sqrt{2-1}}=\displaystyle\frac{1}{2}\).

Ta có \(y^\prime= \left(\cos x\right)^\prime =-\sin x\). Vậy \(y^\prime\left(\displaystyle\frac{\pi}{6}\right)=-\sin \displaystyle\frac{\pi}{6}=-\displaystyle\frac{1}{2}\).

Ta có \(y^\prime= \left(\tan x\right)^\prime =\displaystyle\frac{1}{\cos^2x}\ \left(x \neq \displaystyle\frac{\pi}{2}+k\pi, k \in \mathbb{Z}\right)\). Vậy \(y^\prime\left(\displaystyle\frac{3\pi}{4}\right)= \displaystyle\frac{1}{\cos^2 \left(\frac{3\pi}{4}\right)}=2\).

Ta có \(f'(x)=\cos x\).

Đạo hàm của hàm số trên tại điểm \(x_0=\displaystyle\frac{\pi}{3}\) là

\(f'\left(\displaystyle\frac{\pi}{3}\right)=\cos \displaystyle\frac{\pi}{3}=\displaystyle\frac{1}{2}\).

Ta có \(f'(x)=\displaystyle\frac{1}{\cos^2 x}\) \(\left(x \neq \displaystyle\frac{\pi}{2}+k \pi,\, k \in \mathbb{Z}\right)\).

Đạo hàm của hàm số trên tại điểm \(x_0=\displaystyle\frac{\pi}{4}\) là

\(f'\left(\displaystyle\frac{\pi}{4}\right)=\displaystyle\frac{1}{\cos^2\left(\displaystyle\frac{\pi}{4}\right)}=2\).

Ta có \(f'(x)=-\displaystyle\frac{1}{\sin^2 x}\) \(\left(x \neq k \pi, k \in \mathbb{Z}\right)\).

Đạo hàm của hàm số trên tại điểm \(x_0=\displaystyle\frac{\pi}{2}\) là: \(f'\left(\displaystyle\frac{\pi}{2}\right)=-\displaystyle\frac{1}{\sin^2\left(\displaystyle\frac{\pi}{2}\right)}=-1\).

Ta có \(y'=2-\sin x \Rightarrow y'(\pi)=2\).

Ta có: \(f'(x) = \left(5\cos x - 3\sin x \right)' = 5(\cos x)' - 3(\sin x)' = -5\sin x - 3\cos x\).

Ta có \(y'=-\sin x+\cos x\).

Ta có \(y'= \left(\cos x -\sin x\right)' =-\sin x -\cos x.\)

Ta có: \(y'=\left(2\sin x\right)'-\left(\cos x\right)'=2\cos x+\sin x\).

\(y'=-\sin x-\cos+2. \)

Ta có \(y'=\cos x\) nên \(y'\left(\displaystyle\frac{\pi}{2}\right)=\cos {\displaystyle\frac{\pi}{2}}=0\).

Ta có \(y=\tan x+\sin x\Rightarrow y'=(\tan x+\sin x)'=\displaystyle\frac{1}{\cos^2x}+\cos x\).

Ta có \(y'=\displaystyle\frac{1}{\sin^2 x}+\displaystyle\frac{1}{\cos^2 x}=\displaystyle\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cdot \cos^2 x}=\displaystyle\frac{1}{\sin ^2 x \cdot \cos ^2 x}\).

Ta có \(y'=(2\sin x-3\cos x)'=2\cos x + 3\sin x\).

Tập xác định của hàm số là \(\mathbb{R}\setminus\{k\pi|k\in \mathbb{Z}\}\).

Đạo hàm \(y'=\left(5\cot x-3\right)'=5\cdot (\cot x)'=5\cdot \left(-\displaystyle\frac{1}{\sin^2 x}\right)=-5\left(1+\cot^2 x\right)\).

Ta có \(f'\left(x \right)=\displaystyle\frac{1}{\cos^2 x}\).

\(S=f\left(\displaystyle\frac{\pi}{4} \right)+f'\left(\displaystyle\frac{\pi}{4} \right)=\tan \left(\displaystyle\frac{\pi}{4} \right)+\displaystyle\frac{1}{\cos^2\left(\displaystyle\frac{\pi}{4} \right)}=1+\displaystyle\frac{1}{\left(\displaystyle\frac{\sqrt{2}}{2} \right)^2}=1+2=3\).

\(f'(x)=-\sin x \Rightarrow f'\left(\displaystyle\frac{\pi}{2}\right) = -1\).

Ta có \(\left(\cot x\right)'=-\displaystyle\frac{1}{\sin ^{2} x}\).

Ta có \(y' = \left(3 \sin x-5 \cos x\right)' = 3\cos x +5\sin x\).

\(y'=6\cos3x-2\sin2x\).

\(y'= (\sin 2x )' + \left(\sin \displaystyle\frac{\pi}{3}\right)' - (1)'= (2x)' \cos 2x + 0 - 0 = 2 \cos 2x\).

\(\Rightarrow y'\left(\displaystyle\frac{\pi}{3}\right) =2 \cos \displaystyle\frac{2\pi}{3}=-1.\)

\(f'(x)=\cos 3x\cdot(3x)'=3\cdot\cos 3x\).

Ta có \(y'=\left(\sin {3x}\right)'-\left(\cos {2x}\right)' =3 \cos {3x}+2 \sin {2x}.\)

Áp dụng công thức tính đạo hàm, dùng đạo hàm của hàm hợp dạng \(\sin u\)

Ta có \(y'= \cos (\displaystyle\frac{\pi}{3}-3x) (\displaystyle\frac{\pi}{3}-3x)'=-3\cos (\displaystyle\frac{\pi}{6}-3x)\).

Ta có \(y'=2(\sin 3x)'+(\cos 2x)'=2\cos 3x.(3x)'-\sin 2x.(2x)'=6\cos 3x-2\sin 2x\).

Ta có: \(f'(x)=2\cos2x\Rightarrow f'\left(\displaystyle\frac{\pi}{4}\right)=2\cos\left(2\cdot\displaystyle\frac{\pi}{4}\right)=2\cos\displaystyle\frac{\pi}{2}=0\).

Ta có : \(y'=-2\sin 2x \Rightarrow y'\left(\displaystyle\frac{\pi}{8}\right)=-2\sin \displaystyle\frac{\pi}{4}=-\sqrt{2}\) .

Ta có \(y'=3\cos(3x+2)\).

Ta có \(y'=3\cos3x\).

Ta có \(y'= (\sin 3x-4\cos 2x)' = 3\cos 3x +8\sin 2x.\)

Ta có \(y'=5\cos 5x+6\sin 2x\).

Ta có \(y'=\displaystyle\frac{3}{\cos^{2}3x}\).

Ta có \(y'=2{(\sin 3x)}'+{(\cos 2x)}'=2\cos 3x\cdot(3x)'-\sin 2x\cdot(2x)'=6\cos 3x-2\sin 2x\).

\(y'=(2x)'\cos 2x-(2x)'\sin 2x=2\cos 2x-2\sin 2x\).

Ta có \(f'(x)=1+\tan^2\left(x-\displaystyle\frac{2\pi}{3}\right).

\Rightarrow f'(0)=1+\tan^2\left(0-\displaystyle\frac{2\pi}{3}\right)=1+\left(\sqrt{3}\right)^2=4\).

Ta có \(y'=[\tan (2x+1)]'=\displaystyle\frac{(2x+1)'}{\cos^2 (2x+1)}=\displaystyle\frac{2}{\cos^2 (2x+1)}\).

Ta có \(y'=\cos 2x-\sin x\Rightarrow y'\left(\displaystyle\frac{\pi}{2}\right)=-2\).

Ta có \(y'=-7 \sin 7 x\).

Ta có \(f'(x)=-\left(\displaystyle\frac{x}{4}\right)'\cdot \sin\displaystyle\frac{x}{4}=-\displaystyle\frac{1}{4}\sin\displaystyle\frac{x}{4}\).

Ta có \(f'(x)=\cos x-10\sin 2x\).

Ta có

\begin{eqnarray*}&y'=0&\Leftrightarrow -2\sin 2x+2=0\\&&\Leftrightarrow \sin 2x=1\\&&\Leftrightarrow 2x=\displaystyle\frac{\pi}{2}+k2\pi, k\in\mathbb{Z}\\&&\Leftrightarrow x=\displaystyle\frac{\pi}{4}+k\pi, k\in\mathbb{Z}.\end{eqnarray*}

Ta có \(f'(x)=-2\sin \left(\displaystyle\frac{5\pi}{6}+x\right)\). Do đó

\[f'\left(\displaystyle\frac{\pi}{6}\right)=-2\sin \left(\displaystyle\frac{5\pi}{6}+\displaystyle\frac{\pi}{6}\right)=-2\sin \pi =0.\]

Ta có \(f'(x)=(\sin2x)'+(\cos3x)'=2\cos2x-3\sin3x\).

Ta có \(y=\cos 2x\) nên \(y'=-2\sin 2x\).

Ta có \(y'=-5\sin 5x\Rightarrow y''=-25\cos 5x\).

Ta có \(f'(x)=2\cos 3x \cdot (3x)'+4(-\sin 5x)\cdot (5x)'=6\cos 3x-20\sin 5x\).

Ta có \(y'=\displaystyle\frac{3}{\cos^23x}\).

Ta có \(f'(x) = (1+\tan^2 2x)\cdot (2x)' = 2(1+\tan^2 2x)\Rightarrow f'(0) = 2\).

Ta có \(y'=\displaystyle\frac{1}{5}\cdot (4x)' \cdot \mathrm{e}^{4x}=\displaystyle\frac{4}{5}\mathrm{e}^{4 x}\).

Ta có \(y'=(2x+3)'2^{2x+3}\ln 2=2^{2x+4}\ln 2=2^{2x+2}\cdot 4\cdot \ln 2=2^{2x+2}\ln 16\).

Ta có \(y'=2^{x+1}\ln2(x+1)'=2^{x+1}\ln2\).

Ta có \(f'(x)=\left[2^{x^2+3x+1}\right]'=2^{x^2+3x+1}(2x+3)\ln 2\).

Ta có \(y' = (x^2)' \cdot 2^{x^2} \ln2 =2x \cdot 2^{x^2}\ln 2 = x \cdot 2^{x^2+1}\ln 2\).

Ta có \(f'(x)=(x^2-x)'\cdot 5^{x^2-x}\cdot\ln5 =(2x-1)5^{x^2-x}\ln5\).

Ta có \(\left(2^{3x}\right)'=2^{3x}3\ln 2.\)

Ta có \(y'=-2\cdot 2^{1-2x}\cdot \ln 2 =-2^{2-2x} \cdot \ln 2\)

Áp dụng công thức đạo hàm:

\begin{eqnarray*}&& \left(a^u\right)’=u'\cdot a^u\cdot \ln a \\&\Rightarrow &f'(x)=\left(7^{x^2+6}\right)’ \\&\Rightarrow &f'(x)=\left(x^2+6\right)’7^{x^2+6}\ln 7 \\&\Rightarrow &f'(x)=2x\cdot 7^{x^2+6}\ln 7.\end{eqnarray*}

Ta có \(y'= 2x\cdot 8^{x^2+1} \ln 8 \Leftrightarrow y'= 6x\cdot 8^{x^2+1} \cdot \ln 2\).

Ta có \(y'=\left(3^{x^2+2}\right)'= 3^{x^2+2}\ln3\cdot\left(x^2+2\right)'=2x\cdot 3^{x^2+2}\ln3\).

Ta có \(y=4^{2x}\Rightarrow y'=4^{2x}\cdot \ln 4\cdot \left(2x\right)'=2\cdot 4^{2x}\cdot \ln 4\).

Ta có \(y'=2\cdot 2^{2x}\ln 2=2^{2x+1}\cdot \ln 2\).

Ta có \(y=2^x\Rightarrow y'=2^x\ln 2\).

Ta có: \(\left(5^x\right)^{\prime}=5^x\cdot \ln 5\).

Ta có

\(y'=13^x.\ln 13\).

\(y'=7^{x^2+x-2}\left(2x+1\right) \ln 7\).

Ta có \(y'=15^x\ln15\).

Ta có \(f(x)=\left(\displaystyle\frac{1}{2}\right)^x=2^{-x}\).

Theo công thức tính đạo hàm ta có \(f'(x)=(-x)' 2^{-x} \ln 2 =-\left(\displaystyle\frac{1}{2}\right)^x \ln 2\).

Ta có \(y'=\left(3^{x^{2}-2x}\right)' = (2x-2)\cdot 3^{x^{2}-2x} \cdot \ln 3 = 2(x-1)\cdot 3^{x^{2}-2x} \cdot \ln 3\).

Ta có \(y'=3^{x+1}\ln 3\).

Suy ra \(y'(1)=3^2\ln 3=9\ln 3\).

\(y'=\left(x^2\right)'\cdot \mathrm{e}^{x^2}=2 x e^{x^2}\).

Ta có \(\left({10}^{x} \right)'={10}^{x}\ln10\).

Ta có \(y' = (x^2-3x)' 3^{x^2 - 3x} \ln 3 = (2x - 3) 3^{x^2 - 3x} \ln 3\).

Ta có \(y'=\left(2^{x+1}\right)'=(x+1)'\cdot 2^{x+1}\cdot \ln 2=2^{x+1}\ln 2.\)

\(y'=(x^2+x)'\mathrm{e}^{x^2+x}=(2x+1)\mathrm{e}^{x^2+x}.\)

Tập xác định của hàm số là \(\mathscr D=\mathbb{R}\).

Khi đó \(f'(x)=\left(2^{2x-x^2}\right)'=(2x-x^2)' \cdot 2^{2x-x^2}\cdot \ln 2=(2-2x)\cdot 2^{2x-x^2}\cdot \ln 2=(1-x)\cdot 2^{1+2x-x^2}\cdot \ln 2\).

Ta có \(y'=2\cdot 3^{2x}\ln 3\).

Ta có \(y'=(x^2+2)'\cdot 3^{x^2+2}\cdot\ln 3=2x\cdot 3^{x^2+2}\cdot\ln 3\).

Ta có \(y'=(x^2-3x)'\cdot3^{x^2-3x}\cdot\ln3=(2x-3)\cdot3^{x^2-3x}\cdot\ln3\).

Ta có \(y'=6^x\ln 6\).

Ta có \(f'(x)=\displaystyle\frac{(x^2+2x)'}{(x^2+2x)\ln4}=\displaystyle\frac{2x+2}{2(x^2+2x)\ln2}=\displaystyle\frac{x+1}{(x^2+2x)\ln2}\).

\(y'=\displaystyle\frac{(3\mathrm{e}^x)'}{3\mathrm{e}^x\cdot\ln 2}=\displaystyle\frac{3\mathrm{e}^x}{3\mathrm{e}^x\cdot\ln 2}=\displaystyle\frac{1}{\ln 2}\).

Ta có \(y'=\displaystyle\frac{(1-x^2)'}{1-x^2}=\displaystyle\frac{-2x}{1-x^2}.\)

\(y'=\displaystyle\frac{1}{2x+1}\cdot (2x+1)'=\displaystyle\frac{2}{2x+1}\).

Ta có \(y'=\displaystyle\frac{1}{(6x-5)\ln8}\cdot (6x-5)'=\displaystyle\frac{1}{(6x-5)\cdot 3\cdot\ln2}\cdot 6=\displaystyle\frac{2}{(6x-5)\ln2}\).

Ta có

\(y'=\displaystyle\frac{(4x+2)'}{(4x+2)\ln3}=\displaystyle\frac{4}{(4x+2)\ln3}\).

\(y'=\displaystyle\frac{2x}{\left(x^2+1\right) \ln 2}.\)

Ta có \(y'=\left(\ln(x^2+1)\right)'=\displaystyle\frac{(x^2+1)'}{x^2+1} = \displaystyle\frac{2x}{x^2+1}\).

Ta có \(y'=\displaystyle\frac{(4x-x^2)'}{4x-x^2}=\displaystyle\frac{4-2x}{4x-x^2}=\displaystyle\frac{2(2-x)}{4x-x^2}\).

Ta có \(y'=\displaystyle\frac{2+\mathrm{e}^2}{\left(2x+\mathrm{e}^x\right)\ln2}\).

Ta có \(y'=(2x)'\cdot 7^{2x}\cdot \ln 7-\displaystyle\frac{(5x)'}{5x\ln 2}=2\cdot 7^{2x}\cdot\ln 7 -\displaystyle\frac{1}{x\ln 2}\).

Ta có \(y'=\displaystyle\frac{(x^2+2)'}{(x^2+2) \ln 3} = \displaystyle\frac{2x}{(x^2+2) \ln 3}\).

Ta có \(y'=\left(\log_{7}(x^2+2x+2)\right)'=\displaystyle\frac{(x^2+2x+2)'}{(x^2+2x+2)\cdot\ln{7}}=\displaystyle\frac{2x+2}{(x^2+2x+2)\cdot\ln{7}}\).

\(y'=\displaystyle\frac{3^x\ln 3+1}{(3^x+x)\ln 3}\), \(y'(1)=\displaystyle\frac{3\ln 3 +1}{4 \ln 3}=\displaystyle\frac{3}{4}+\displaystyle\frac{1}{4 \ln 3}\). Ta có \(a=3\), \(b=4\). Vậy \(a+b=7\).

\(y'=\displaystyle\frac{2x}{(x^2+1)\ln 2}\).

Ta có \(y'=\displaystyle\frac{(x^2-2x+1)'}{(x^2-2x+1)\ln 3}=\displaystyle\frac{2x-2}{(x-1)^2\ln 3}=\displaystyle\frac{2}{(x-1)\ln 3}\).

Tập xác định của hàm số: \(\mathscr{D}=\left(-\displaystyle\frac{2}{3};+\infty\right)\).

Ta có \(y=\log _3\left(3x+2 \right) \Rightarrow y'=\displaystyle\frac{3}{\left(3x+2 \right)\ln3}\).

Ta có \(y'=\mathrm{e}^x -\displaystyle\frac{(3x)'}{3x}=\mathrm{e}^x-\displaystyle\frac{3}{3x}=\mathrm{e}^x-\displaystyle\frac{1}{x}.\)

\(y'= \displaystyle\frac{(x^2-x+1)'}{(x^2-x+1) \ln 10}=\displaystyle\frac{2x - 1}{\left(x^2 - x + 1\right)\ln 10}.\)

Ta có \(y'=\displaystyle\frac{(x+\mathrm{e}^x)'}{(x+\mathrm{e}^x)\ln2}=\displaystyle\frac{1+\mathrm{e}^x}{(x+\mathrm{e}^x)\ln2}\).

Ta có \(f'(x)=1+\displaystyle\frac{1}{x+3}\).

Đạo hàm của hàm số đã cho là

\(y'=\displaystyle\frac{\left(x^2+1\right)'}{\left(x^2+1\right)\ln5}=\displaystyle\frac{2x}{\left(x^2+1\right)\ln5}\).

Ta có \(f'(x)=\left[\log_3\left(x^2+x\right)\right]'=\displaystyle\frac{\left(x^2+x\right)'}{\left(x^2+x\right)\ln 3}=\displaystyle\frac{2x+1}{\left(x^2+x\right)\ln 3}\).

\(y'=\displaystyle\frac{2^x\ln 2}{(2^x+1)\ln 2}=\displaystyle\frac{2^x}{2^x+1}\).

\(f'(x)=\displaystyle\frac{1}{(x^2+1)\ln 3} \cdot (x^2+1)'=\displaystyle\frac{2x}{ (x^2+1)\ln 3 }\).

Ta có \(y'=\left[\ln(x^2-3x+2)\right]'=\displaystyle\frac{(x^2-3x+2)'}{x^2-3x+2}=\displaystyle\frac{2x-3}{x^2-3x+2}\).

Ta có \(y'=\displaystyle\frac{1}{x\ln 5}\).

Ta có \(y' = \displaystyle\frac{(4x + 1)'}{(4x + 1) \ln 3} = \displaystyle\frac{4}{(4x + 1) \ln 3}\).

Ta có \(y'=\displaystyle\frac{(x+1)'}{x+1}=\displaystyle\frac{1}{x+1}\).

Đạo hàm của hàm số \(f(x)=\log_2 x\) là \(f'(x)=\displaystyle\frac{1}{x\ln2}\).

Ta có \(y'=\displaystyle\frac{\left(x^2+1\right)^\prime}{\left(x^2+1\right)\ln9}=\displaystyle\frac{2x}{\left(x^2+1\right)\ln{3^2}}=\displaystyle\frac{2x}{\left(x^2+1\right)2\ln3}=\displaystyle\frac{x}{\left(x^2+1\right)\ln 3}\).

Ta có \(y' = \left[\ln \left(x^{2}+x+1\right)\right]'= \displaystyle\frac{\left(x^{2}+x+1\right)'}{x^{2}+x+1} = \displaystyle\frac{2x+1}{x^{2}+x+1}\).

\(y'=\displaystyle\frac{(1-5x)'}{(1-5x)\ln 3}=-\displaystyle\frac{5}{(1-5x)\ln 3}\).

Ta có \(y'=\displaystyle\frac{(x^2-3x-4)'}{(x^2-3x-4)\ln 8}=\displaystyle\frac{2x-3}{(x^2-3x-4)\ln 8}.\)

\(y'=\displaystyle\frac{1}{(x-1)\ln 2}\).

Ta có \(y'=\left[\ln (x^4+4x^3-3)\right]'=\displaystyle\frac{(x^4+4x^3-3)'}{x^4+4x^3-3}=\displaystyle\frac{4x^3+12x^2}{x^4+4x^3-3}\).

Ta có \(y'=\displaystyle\frac{(5x-3)'}{(5x-3)\ln 2}=\displaystyle\frac{5}{(5x-3)\ln 2}\Rightarrow\begin{cases} a=5\\b=2\end{cases}\Rightarrow a+b=7\).

Ta có \(f'(x)=\displaystyle\frac{\left(2x+1\right)'}{\left(2x+1\right)\ln 3}=\displaystyle\frac{2}{\left(2x+1\right)\ln 3}\Rightarrow f'(0)=\displaystyle\frac{2}{\ln 3}\).

Ta có \(y'=\displaystyle\frac{(x+\mathrm{e}^{x})'}{(x+\mathrm{e}^{x})\cdot \ln 2}=\displaystyle\frac{1+\mathrm{e}^{x}}{(x+\mathrm{e}^{x})\cdot \ln 2}\).

\(y=\log_2(2x^2+x+3)\Rightarrow y'=\displaystyle\frac{{(2x^2+x+3)}'}{(2x^2+x+3)\cdot \ln 2}=\displaystyle\frac{4x+1}{(2x^2+x+3)\cdot \ln 2}.\)

Ta có \(y'=\displaystyle\frac{(2x+1)'}{(2x+1) \ln 3}=\displaystyle\frac{2}{(2x+1) \ln 3}\).

Ta có: \(f'(x)=2(x^3-2x^2)'(x^3-2x^2)=2(3x^2-4x)(x^3-2x^2)=6x^5-20x^4+16x^3\).

Ta có

\(y'=4(7x-5)'(7x-5)^3=28(7x-5)^3=-28(5-7x)^3.\)

Ta có: \(y'=5(1-x^3)'(1-x^3)^4=5(-3x^2)(1-x^3)^4=-15x^2(1-x^3)^4.\)

Ta có \(y'=20(x^3-2x^2)(x^3-2x^2)^{19}=20(3x^2-4x)(x^3-2x^2)^{19}\).

Ta có: \(y'=\left[(7x-5)^{4}\right]'=4.7(7x-5)^3= 28(7x-5)^{3}\).

Ta có \(y' = \left[ (x^3 - 2x^2)^2 \right]' = 2(x^3 - 2x^2)(x^3 - 2x^2)' = 2(x^3 - 2x^2)(3x^2 - 4x) \Rightarrow y'(1) = 2\).

Ta có \(y'=4(x^2+2x+3)^3(2x+2)\).

Ta có \(f'(x)=8x(2x^2-1) \Rightarrow f'(\sqrt{2})=24\sqrt{2}\).

}

Ta có \(y' = 3(x^2+3)' (x^2+3)^2 = 6x(x^2+3)^2.\)

Ta có \(y'=4(5x-1)^3\cdot (5x-1)'=20(5x-1)^3\).

Ta có \(y'=11(4x^4-3x)^{10}(4x^4-3x)'=11(16x^3-3)(4x^4-3x)^{10}\).

Ta có \(y'=2\left(x^3-2x^2 \right)(3x^2-4x)=6x^5-20x^4+16x^3\).

Ta có \([(x^2-3x)^5]'=5(2x-3)(x^2-3x)^4\).

Ta có

\begin{eqnarray*}\left[\left(x^2-4x+5\right)^{\sqrt{3}}\right]'&=&\sqrt{3}\cdot \left(x^2-4x+5\right)^{\sqrt{3}-1}\cdot(x^2-4x+5)'\\ &=&\sqrt{3}(2x-4)\left(x^2-4x+5\right)^{\sqrt{3}-1}.\end{eqnarray*}

\(y'=(\sin^2x)'=2\sin x \cos x =\sin 2x\).

Ta có \(y'= 2\cos (2x) \cdot (\cos 2x)'=2\cos (2x)\cdot (-2)\sin (2x) =-2\sin (4x)\).

Ta có \(y'=2\left(\cos x\right)'\left(\cos x\right)=-2\sin x\cos x=-\sin 2x\).

Ta có: \(y'=\displaystyle\frac{1}{2\sqrt{\cot x}}(\cot x)'=-\displaystyle\frac{1}{2{\sin^2}x\sqrt{\cot x}}\).

\(f'(x)=(\cos2x)'-(\sin^2x)'=-2\sin2x-2\sin x\cos x=-2\sin2x-\sin2x=-3\sin2x\).

Ta có \(y'=-2\cot x\cdot(\cot x)'=-2\cot x \cdot \left(-(1+\cot^2x)\right)=2\cot x(1+\cot^2x).\)

Ta có \(y'=2\cdot\sin2x\cdot\left(\sin2x\right)'=2\cdot\sin2x\cdot\left(2x\right)'\cos2x=2\cdot2\cdot\sin2x\cdot\cos2x=2\sin4x\).

\(y'=\displaystyle\frac{(\sin 3x)'}{2\sqrt{\sin 3x}}=\displaystyle\frac{3\cos 3x}{2\sqrt{\sin 2x}}.\)

Ta có: \(y'=\displaystyle\frac{2}{\cos^2(2x)}+2\cot x\left(\displaystyle\frac{-1}{\sin^2x}\right)=\displaystyle\frac{2}{\cos^2(2x)}-\displaystyle\frac{2\cot x}{\sin^2x}\).

Hàm số \(f(x)=\sin ^2 2 x-\cos 3 x\) có đạo hàm là

\[f'(x)=2\cdot2\cdot \sin2x \cdot \cos2x+3\sin3x=2\sin4x+3\sin3x.\]

Ta có \(y' = 2 \cos 3x \cdot \left(\cos 3x\right)' = 2 \cos 3x \cdot\left(-3 \sin 3x\right) = -3 \sin 6x\).

Ta có \(y'=2 \tan x\cdot (\tan x)'=2 \tan x\cdot \displaystyle\frac{1}{\cos^2 x}=\displaystyle\frac{2\tan x}{\cos ^2 x}\).

Ta có \(y=x+\sin ^{2} x\Rightarrow y'=1+2\cos x\sin x=1+\sin 2x\).

Ta có \(y'=(x)'\cdot\cos2x+x\cdot(\cos2x)'=\cos2x+x\cdot(-2)\cdot\sin 2x=\cos2x-2x\sin2x\).

Ta có

\begin{align*}y'=\ &(x-2)'\sqrt{x^2+3}+(x-2){\left(\sqrt{x^2+3}\right)}'\\ =\ &\sqrt{x^2+3}+(x-2) \cdot \displaystyle\frac{x}{\sqrt{x^2+3}}\\ =\ &\displaystyle\frac{2x^2-2x+3}{\sqrt{x^2+3}}.\end{align*}

Ta có

\begin{eqnarray*}y'&=& (x^2-1)'(x^2-4)(x^2-9)+(x^2-1)(x^2-4)'(x^2-9)+(x^2-1)(x^2-4)(x^2-9)'\\& =&2x(x^2-4)(x^2-9)+(x^2-1)2x(x^2-9)+(x^2-1)(x^2-4)2x\\& =&2x(3x^4-28x^2+49). \end{eqnarray*}

Ta có \(y'=\sin x+x\cos x.\)

Ta có \(y'=2x\cos x-x^2\sin x\).

Ta có \(f'(x)=\left(x\cos x \right)'= \cos x -x\sin x\).

Suy ra, \(f'\left(\displaystyle\frac{\pi}{6}\right) = \cos \displaystyle\frac{\pi}{6}- \displaystyle\frac{\pi}{6}\sin \displaystyle\frac{\pi}{6}=\displaystyle\frac{\sqrt{3}}{2}-\displaystyle\frac{\pi}{6}\cdot \displaystyle\frac{1}{2}=\displaystyle\frac{6\sqrt{3}-\pi}{12}.\)

Ta có \(y'=\sin x+x\cos x-\sin x=x\cos x\).

\(y=x\tan 2x \Rightarrow y'=(x))\tan 2x+x{\left(\tan 2x\right)}'\)

\(=\tan 2x+x\displaystyle\frac{(2x)'}{\cos^22x}=\tan 2x+\displaystyle\frac{2x}{\cos^22x}=2x\tan^22x+\tan 2x+2x\).

Ta có \(y'=\cot x-\displaystyle\frac{x}{\sin^{2}x}\).

Ta có \(y'=(\sin x)'- (x \cos x)'=\cos x -(\cos x -x\sin x)=x\sin x\).

Ta có \(y'=(x^3)'\cdot \cos x+x^3\cdot (\cos x)'=3x^2\cos x-x^3\sin x\).

\(y=2\sin 3x \cdot \cos 5x=\sin 8x - \sin 2x\)

\(y'=(\sin 8x)' - (\sin 2x)'=8\cos 8x - 2\cos 2x=2(4\cos 8x-\cos 2x)\).

Ta có: \(y' = (x)'.\cot 2x + (\cot 2x)'.x = \cot 2x - 2x(1 + \cot^2 2x) = -2x \cot^2 2x + \cot 2x - 2x\).

Ta có \(y'=\left(\mathrm{e}^x\right) '\sin 2x+\left(\sin 2x\right) '\mathrm{e}^x=\mathrm{e}^x\left(\sin 2x+2\cos 2x\right) .\)

Ta có: \(f'(x)=\ln x+1\), suy ra \(P=f(x)-x\cdot f'(x)+x=x\ln x-x(\ln x+1)+x=0\).

Ta có \(y'=2^x\ln 2\cdot \ln x+2^x\cdot \displaystyle\frac{1}{x}=2^x\left(\ln 2\cdot \ln x+\displaystyle\frac{1}{x}\right).\)

Ta có \(f'(x)=\mathrm{e}^{-2x}-2x\mathrm{e}^{-2x}\) và \(f''(x)=-2\mathrm{e}^{-2x}-2\left(\mathrm{e}^{-2x}-2x\mathrm{e}^{-2x}\right)=-4\mathrm{e}^{-2x}+4x\mathrm{e}^{-2x}\).

Do đó \(f''\left(-\displaystyle\frac{1}{2}\right)=-6\mathrm{e}\).

\(y=(x^2-x+1)\mathrm{e}^x\Rightarrow y'=(2x-1)\mathrm{e}^x+(x^2-x+1)\mathrm{e}^x=(x^2+x)\mathrm{e}^x\).

\(y'=\mathrm{e}^x+x\mathrm{e}^x=(1+x)\mathrm{e}^x\).

\(y'=\mathrm{e}^x (x^2+mx) +\mathrm{e}^x (2x+m) =\mathrm{e}^x (x^2+(m+2)x+m)\).

\(y'(0)=1 \Leftrightarrow m=1 \Rightarrow y'=\mathrm{e}^x (x^2+3x+1)\).

Vậy \(y'(1)=5\mathrm{e}\).

\(y'=(x+2)' e^{2x} +(x+2)\left(e^{2x} \right)=e^{2x} +2(x+2)e^{2x} =(2x+5)e^{2x} \).

Ta có \(y'=\mathrm{e}^{x+1}+x\mathrm{e}^{x+1}=(x+1)\mathrm{e}^{x+1}\).

\(y' = (2x-3)\mathrm{e}^x + (x ^2 -3x+3)\mathrm{e}^x = (x^2-x)\mathrm{e}^x\).

\(y'=(x2^x)'=2^x+x2^x\ln2=2^x(1+x\ln 2).\)

\(y^\prime =\sqrt{x^2+1}+(x+1)\cdot \displaystyle\frac{x}{\sqrt{x^2+1}}=\displaystyle\frac{x^2+1+x^2+x}{\sqrt{x^2+1}}=\displaystyle\frac{2x^2+x+1}{\sqrt{x^2+1}}\).

Có \(y=(1+x)\sqrt{1-x}\Rightarrow y'=1\cdot\sqrt{1-x}+\displaystyle\frac{-1}{2\sqrt{1-x}}\cdot(1+x)=\displaystyle\frac{-3x+1}{2\sqrt{1-x}}\).

Suy ra \(a=-3\), \(b=1\).

Vậy \(a+2b=-1\).

Ta có \(f'(x)=\sqrt{x^2+1}+\displaystyle\frac{x^2}{\sqrt{x^2+1}}=\displaystyle\frac{2x^2+1}{\sqrt{x^2+1}}\).

Suy ra \(a=2\), \(b=0\), \(c=1\). Suy ra \(a^2+b^3+3c^2=7\).

Ta có

\[f'(x)=\sqrt{x^2+1}+x\cdot \displaystyle\frac{x}{\sqrt{x^2+1}}=\displaystyle\frac{2x^2+1}{\sqrt{x^2+1}}.\]

Vậy \(f'(-2)=\displaystyle\frac{9\sqrt{5}}{5}\).

Ta có \(y'=\sqrt{2-x}-\displaystyle\frac{x+1}{2\sqrt{2-x}}=\displaystyle\frac{2(2-x)-(x+1)}{2\sqrt{2-x}}=\displaystyle\frac{-\displaystyle\frac{3}{2}x+\displaystyle\frac{3}{2}}{\sqrt{2-x}}\).

Suy ra \(\begin{cases} a=-\displaystyle\frac{3}{2}\\b=\displaystyle\frac{3}{2}\end{cases}\Rightarrow 2a+4b=3\).

Ta có

\begin{eqnarray*}f'(x)=\left[ (x-2) \sqrt{x^{2}+1}\right]'&=&(x-2)'\cdot \sqrt{x^{2}+1}+\left(\sqrt{x^{2}+1}\right)'\cdot(x-2)\\&=&\sqrt{x^{2}+1}+\displaystyle\frac{x}{\sqrt{x^2+1}}\cdot(x-2)\\&=&\displaystyle\frac{2x^2-2x+1}{\sqrt{x^2+1}}.\end{eqnarray*}

Từ đó xác định được \(a=2\), \(b=-2\), \(c=1\) suy ra \(S=a-b+c=2-(-2)+1=5\).

Ta có \(y'=\sqrt{x^2+2}+x\cdot\displaystyle\frac{2x}{2\sqrt{x^2+2}}=\displaystyle\frac{2(x^2+2)+2x^2}{2\sqrt{x^2+2}}=\displaystyle\frac{2x^2+2}{\sqrt{x^2+2}}\).

Ta có \(y'=\displaystyle\frac{1}{2\sqrt{x+1}}\ln x+\displaystyle\frac{1}{x}\sqrt{x+1}=\displaystyle\frac{x\ln x+2(x+1)}{2x\sqrt{x+1}}\).

Ta có \(y'=\sqrt{x^2-2x}+x.\displaystyle\frac{2x-2}{2\sqrt{x^2-2x}}=\displaystyle\frac{x^2-2x+x^2-x}{\sqrt{x^2-2x}}=\displaystyle\frac{2x^2-3x}{\sqrt{x^2-2x}}.\)

Ta có

\(y'=(2x-1)'\sqrt{x^2+x}+(2x-1)(\sqrt{x^2+x})'

=2\sqrt{x^2+x}+\displaystyle\frac{(2x-1)(2x+1)}{2\sqrt{x^2+x}}=2\sqrt{x^2+x}+\displaystyle\frac{4x^2-1}{2\sqrt{x^2+x}}.\)

\(y'=\left(\displaystyle\frac{x+1}{4^x}\right)'=\displaystyle\frac{1\cdot 4^x -4^x (x+1)\ln 4}{4^{2x}}=\displaystyle\frac{1 - (x+1)\ln 4}{4^{x}}=\displaystyle\frac{1-2(x+1)\ln 2}{2^{2x}}.\)

Tập xác định của hàm số \(\mathscr{D}=(0;+\infty)\). Ta có

\[f'(x)=\displaystyle\frac{\left(\log_2x \right)'x-\left(\log_2x \right)\cdot(x)' }{x^2}=\displaystyle\frac{\displaystyle\frac{1}{x\ln2}\cdot x-\log_2x}{x^2}=\displaystyle\frac{\displaystyle\frac{1}{\ln2}-\log_2x}{x^2}=\displaystyle\frac{1-\ln x}{x^2\ln2}.\]

Ta có \(y'=\left(\displaystyle\frac{x+3}{\sqrt{x^{2}+1}}\right)'=\displaystyle\frac{(x+3)'(\sqrt{x^{2}+1})-(x+3)(\sqrt{x^{2}+1})'}{(\sqrt{x^{2}+1})^2}=\displaystyle\frac{\sqrt{x^{2}+1}-(x+3)\displaystyle\frac{x}{\sqrt{x^{2}+1}}}{x^{2}+1}=\displaystyle\frac{1-3 x}{\left(x^{2}+1\right) \sqrt{x^{2}+1}}\).

\(y=\displaystyle\frac{x}{\cos x}\Rightarrow y'=\displaystyle\frac{x'\cos x-x(\cos x)'}{\cos^2x}=\displaystyle\frac{\cos x+x\sin x}{\cos^2x}\).

Ta có

\begin{eqnarray*}y' =\left(\displaystyle\frac{\sqrt{x-1}}{x}\right)'&=&\displaystyle\frac{x\left(\sqrt{x-1}\right)'-\sqrt{x-1} \cdot (x)'}{x^2}\\&=&\displaystyle\frac{\displaystyle\frac{x}{2\sqrt{x-1}}-\sqrt{x-1}}{x^2}\\&=&\displaystyle\frac{x-2(x-1)}{2x^2\sqrt{x-1}}\\&=&\displaystyle\frac{2-x}{2x^2\sqrt{x-1}}.\end{eqnarray*}

Ta có \(y'=\displaystyle\frac{2\sqrt{x^2+1}-(2x+3)\displaystyle\frac{x}{\sqrt{x^2+1}}}{x^2+1} =\displaystyle\frac{2-3x}{(x^2+1)\cdot \sqrt{x^2+1}}\).

Ta có \(y'=\left(\displaystyle\frac{1}{\sqrt{x^2+1}}\right)'=\displaystyle\frac{-(\sqrt{x^2+1})'}{x^2+1}=\displaystyle\frac{-(x^2+1)'}{2\sqrt{x^2+1}(x^2+1)}=\displaystyle\frac{-x}{\sqrt{x^2+1}(x^2+1)}.\)

Ta có \(f'(x)=\displaystyle\frac{x'\sqrt{4-x^2}-x(\sqrt{4-x^2})'}{4-x^2}=\displaystyle\frac{\sqrt{4-x^2}-x\cdot \displaystyle\frac{-x}{\sqrt{4-x^2}}}{4-x^2}=\displaystyle\frac{4-x^2+x^2}{\sqrt{4-x^2}(4-x^2)}=\displaystyle\frac{4}{(4-x^2)\sqrt{4-x^2}}.\)

Suy ra \(y'(0)=\displaystyle\frac{4}{4\sqrt{4}}=\displaystyle\frac{1}{2}.\)

Ta có

\begin{eqnarray*}y'&=&\displaystyle\frac{(x-1)'\sqrt{x^2+1}-(x-1)(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}\\&=&\displaystyle\frac{\sqrt{x^2+1}-(x-1)\displaystyle\frac{x}{\sqrt{x^2+1}}}{(\sqrt{x^2+1})^2}\\&=&\displaystyle\frac{x^2+1-x^2+x}{(\sqrt{x^2+1})^3}\\&=&\displaystyle\frac{1+x}{\sqrt{(x^2+1)^3}}.\end{eqnarray*}

Ta có \(y'=\displaystyle\frac{-\sin x\left(1-\sin x+\right) \cos x\cdot\cos x}{\left(1-\sin x\right)^2 }= \displaystyle\frac{-\sin x+\sin^2 x+\cos^2 x}{\left(1-\sin x\right)^2 }= \displaystyle\frac{-\sin x+1}{\left(1-\sin x\right)^2 }=\displaystyle\frac{1}{ 1-\sin x }.\)

Khi đó \(y'\left(\displaystyle\frac{\pi}{6}\right)=\displaystyle\frac{1}{ 1-\displaystyle\frac{1}{2} }=2\).

Ta có \(f'(x)=\displaystyle\frac{\cos x(1+\cos x)-(-\sin x)\sin x}{(1+\cos x)^2}=\displaystyle\frac{\cos x+\cos^2x+\sin^2x}{(1+\cos x)^2}=\displaystyle\frac{1+\cos x}{(1+\cos x)^2}=\displaystyle\frac{1}{1+\cos x}\).

Do đó \(a=b=1\). Vậy \(a+b=2\).

Ta có

\begin{eqnarray*}f'(x)&=&\displaystyle\frac{\left(3^x-1\right)'\left(3^x+1\right)-\left(3^x-1\right)\left(3^x+1\right)'}{\left(3^x+1\right)^2}\\&=&\displaystyle\frac{\left(3^x+1\right)3^x\ln 3-\left(3^x-1\right)3^x\ln 3}{\left(3^x+1\right)^2}\\&=&\displaystyle\frac{2}{\left(3^x+1\right)^2}\cdot 3^x\ln 3.\end{eqnarray*}

Vậy \(f'(x)=\displaystyle\frac{2}{\left(3^x+1\right)^2}\cdot 3^x\ln 3\).

Ta có \(f(x)=\displaystyle\frac{9}{6^{x}}=9\cdot 6^{-x}\), suy ra \(f'(x)=-9\cdot 6^{-x}\ln 6\).

Ta có

\begin{eqnarray*}y'&=&\displaystyle\frac{(\sin x-\cos x)'(\sin x+\cos x)-(\sin x-\cos x)(\sin x+\cos x)'}{(\sin x+\cos x)^2}\\&=&\displaystyle\frac{(\cos x+\sin x)(\sin x+\cos x)-(\sin x-\cos x)(\cos x-\sin x)'}{1+2\sin x\cos x}\\&=&\displaystyle\frac{(\sin x+\cos x)^2+(\sin x-\cos x)^2}{1+\sin 2x}\\&=& \displaystyle\frac{1+\sin 2x+1-\sin 2x}{1+\sin 2x}\\&=&\displaystyle\frac{2}{1+\sin 2x}.\end{eqnarray*}

Ta có \(y=\displaystyle\frac{\sin x+\cos x}{\sin x-\cos x}=\displaystyle\frac{\sqrt{2}\sin \left(x+\displaystyle\frac{\pi}{4}\right)}{-\sqrt{2}\cos \left(x+\displaystyle\frac{\pi}{4}\right)}=-\tan \left(x+\displaystyle\frac{\pi}{4}\right).\)

Suy ra \(y'=-\displaystyle\frac{1}{\cos^2\left(x+\displaystyle\frac{\pi}{4}\right)}=-\displaystyle\frac{1}{\left(\displaystyle\frac{\cos x-\sin x}{\sqrt{2}}\right)^2}=\displaystyle\frac{-2}{(\sin x-\cos x)^2}.\)

Ta có \(y'=-\displaystyle\frac{-2(\tan (1-2x))'}{\tan^2(1-2x)}=\displaystyle\frac{-4\cdot \displaystyle\frac{1}{\cos^2(1-2x)}}{\tan^2(1-2x)}=\displaystyle\frac{-4}{\sin^2(1-2x)}.\)

Ta có \(y'=\displaystyle\frac{(\cos 2x)'(3x+1)-(3x+1)'\cos2x}{(3x+1)^2}=\displaystyle\frac{-2(3x+1)\sin 2x-3\cos 2x}{(3x+1)^2}.\)

Với điều kiện \(x>\displaystyle\frac{1}{4}\) ta có

\begin{align*}\left(\displaystyle\frac{2-2x}{\sqrt{4x-1}}\right)'&=\displaystyle\frac{-2\sqrt{4x-1}-(2-2x)\cdot\displaystyle\frac{4}{2\sqrt{4x-1}}}{4x-1}\\&=\displaystyle\frac{-2(4x-1)-2(2-2x)}{(4x-1)\sqrt{4x-1}}=\displaystyle\frac{-4x-2}{(4x-1)\sqrt{4x-1}}.\end{align*}

Suy ra \(a=-4\), \(b=2\) và \(E=-2\).

Ta có \(y'=\displaystyle\frac{1}{2\sqrt{\displaystyle\frac{2x-1}{x+2}}} \cdot \left(\displaystyle\frac{2x-1}{x+2}\right)'=\displaystyle\frac{1}{2}\cdot \displaystyle\frac{5}{(x+2)^2} \sqrt{\displaystyle\frac{x+2}{2x-1}}.\)

Ta có \(y'=\displaystyle\frac{1}{2\sqrt{\displaystyle\frac{x^2+1}{x}}}\left(\displaystyle\frac{x^2+1}{x}\right)'=\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{x}{x^2+1}}\left(1-\displaystyle\frac{1}{x^2}\right).\)

\(y' = \displaystyle\frac{1}{2\sqrt{2 + \cos^2 2x}}\cdot \left(2 + \cos^2 2x \right)'= \displaystyle\frac{1}{2\sqrt{2 + \cos^2 2x}}\cdot 2 \cos 2x \cdot (\cos 2x)'\\= \displaystyle\frac{\cos 2x}{\sqrt{2 + \cos^2 2x}}\cdot (-\sin 2x) \cdot (2x)' = \displaystyle\frac{- \sin 4x}{\sqrt{2 + \cos^2 2x}}.\)

\(\begin{aligned}y= \cos\sqrt{x^2+1}\Rightarrow y'=&-\left(\sqrt{x^2+1}\right)'\sin \sqrt{x^2+1}=-\displaystyle\frac{\left(x^2+1\right)'}{2\sqrt{x^2+1}}\sin \sqrt{x^2+1}\\ =&-\displaystyle\frac{x}{\sqrt{x^2+1}}\sin\sqrt{x^2+1}.\end{aligned}\)

Ta có \(f'(x)=\displaystyle\frac{1}{2\sqrt{x}}\cos \sqrt{x} + \displaystyle\frac{1}{2\sqrt{x}}\sin\sqrt{x}\).

Do đó \(f'\left(\displaystyle\frac{\pi^2}{16}\right) =\displaystyle\frac{2}{\pi}\cos\displaystyle\frac{\pi}{4} +\displaystyle\frac{2}{\pi}\sin\displaystyle\frac{\pi}{4} =\displaystyle\frac{2\sqrt{2}}{\pi}\).

\(y' = \cos \sqrt{2 + x^2} \cdot (\sqrt{2 + x^2})' = \cos \sqrt{2 + x^2} \cdot \displaystyle\frac{x}{\sqrt{2+ x^2}}\).

Ta có\(\left[f(x)\right]'=\left[\log_7\left(x\mathrm{e}^x\right)\right]'=\displaystyle\frac{\left(x\mathrm{e}^x\right)'}{x\mathrm{e}^x\ln 7}=\displaystyle\frac{\mathrm{e}^x+x\mathrm{e}^x}{x\mathrm{e}^x\ln 7}=\displaystyle\frac{\mathrm{e}^x\left(1+x\right)}{x\mathrm{e}^x\ln 7}=\displaystyle\frac{1+x}{x\ln 7}\).

Ta có \(f'(x) = \left(\ln \displaystyle\frac{x + 1}{x - 1}\right)' = \displaystyle\frac{\left(\frac{x + 1}{x - 1}\right)'}{\frac{x + 1}{x - 1}} = \displaystyle\frac{\frac{-2}{(x - 1)^2}}{\frac{x + 1}{x - 1}} = \displaystyle\frac{-2}{x^2 - 1}\).

Hàm số đã cho xác định khi \(x\geq -1\).

Với mọi \(x>-1\) ta có

\[y' = \displaystyle\frac{\left(1+\sqrt{x+1}\right)'}{\left(1+\sqrt{x+1}\right)\ln 10} = \displaystyle\frac{1}{2\sqrt{x+1}\left(1+\sqrt{x+1}\right)\ln 10}.\]

\(f'(x) = \mathrm{e}^x \cdot \mathrm{e}^{\mathrm{e}^x} \Rightarrow f'(1)=\mathrm{e}^{\mathrm{e}+1}\).

\(y=\mathrm{e}^{\sin x}\Rightarrow y'=\left(\sin x\right)'\cdot \mathrm{e}^{\sin x}=\cos x\cdot\mathrm{e}^{\sin x}\).

Ta có: \(f'(x)=\displaystyle\frac{1}{2\sqrt{\ln(\ln x)}}{(\ln(\ln x))'}=\displaystyle\frac {1}{2\sqrt{\ln(\ln x)}}\cdot\displaystyle\frac{1}{\ln x}{(\ln x)'}=\displaystyle\frac {1}{2x\ln x\sqrt{\ln(\ln x)}}\).

Vậy \(f'(x)=\displaystyle\frac{1}{2x\ln x\sqrt{\ln(\ln x)}}\).

Ta có \(y'=\displaystyle\frac{\left(\sqrt{x}+1\right)'}{\sqrt{x}+1}=\displaystyle\frac{\displaystyle\frac{1}{2\sqrt{x}}}{\sqrt{x}+1}=\displaystyle\frac{1}{2\sqrt{x}\left(\sqrt{x}+1\right)}=\displaystyle\frac{1}{2x+2\sqrt{x}}\).