a: \(y'=\left(x^2+2x\right)'\left(x^3-3x\right)+\left(x^2+2x\right)\left(x^3-3x\right)'\)

\(=\left(2x+2\right)\left(x^3-3x\right)+\left(x^2+2x\right)\left(3x^2-3\right)\)

\(=2x^4-6x^2+2x^3-6x+3x^4-3x^2+6x^3-6x\)

\(=5x^4+8x^3-9x^2-12x\)

b: y=1/-2x+5 

=>\(y'=\dfrac{2}{\left(2x+5\right)^2}\)

c: \(y'=\dfrac{\left(4x+5\right)'}{2\sqrt{4x+5}}=\dfrac{4}{2\sqrt{4x+5}}=\dfrac{2}{\sqrt{4x+5}}\)

d: \(y'=\left(sinx\right)'\cdot cosx+\left(sinx\right)\cdot\left(cosx\right)'\)

\(=cos^2x-sin^2x=cos2x\)

e: \(y=x\cdot e^x\)

=>\(y'=e^x+x\cdot e^x\)

f: \(y=ln^2x\)

=>\(y'=\dfrac{\left(-1\right)}{x^2}=-\dfrac{1}{x^2}\)

a: TXĐ: \(D=R\backslash\left\{-\dfrac{1}{2}\right\}\)

b: TXĐ: \(D=R\backslash\left\{-3;1\right\}\)

c: TXĐ: \(D=\left[-\dfrac{1}{2};3\right]\)

Lời giải:
a. Hệ số 2>0 nên hàm đồng biến 

b. Hệ số $1-\sqrt{2}<0$ nên hàm nghịch biến 

c. Hệ số $-5<0$ nên hàm nghịch biến 

d. Hệ số $1+m^2>0$ với mọi $m\in\mathbb{R}$ nên hàm đồng biến

e. Hệ số $\sqrt{3}-1>0$ nên hàm đồng biến 

f. Hệ số $2+m^2>0$ với mọi $m\in\mathbb{R}$ nên hàm đồng biến.

e: \(f\left(-x\right)=\dfrac{\left(-x\right)^4+3\cdot\left(-x\right)^2-1}{\left(-x\right)^2-4}=\dfrac{x^4+3x^2-1}{x^2-4}=f\left(x\right)\)

Vậy: f(x) là hàm số chẵn

\(c,f\left(-x\right)=\sqrt{-2x+9}=-f\left(x\right)\)

Vậy hàm số lẻ

\(d,f\left(-x\right)=\left(-x-1\right)^{2010}+\left(1-x\right)^{2010}\\ =\left[-\left(x+1\right)\right]^{2010}+\left(x-1\right)^{2010}\\ =\left(x+1\right)^{2010}+\left(x-1\right)^{2010}=f\left(x\right)\)

Vậy hàm số chẵn

\(g,f\left(-x\right)=\sqrt[3]{-5x-3}+\sqrt[3]{-5x+3}\\ =-\sqrt[3]{5x+3}-\sqrt[3]{5x-3}=-f\left(x\right)\)

Vậy hàm số lẻ

\(h,f\left(-x\right)=\sqrt{3-x}-\sqrt{3+x}=-f\left(x\right)\)

Vậy hàm số lẻ

a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)

=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)

\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)

b: \(y=\left(3x+1\right)^{\Omega}\)

=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)

=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)

c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)

=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)

\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)

\(=\dfrac{-x}{\left(x-1\right)^2}\cdot\dfrac{\sqrt[3]{\left(x-1\right)^2}}{3}\)

\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)

d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)

\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)

\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)

\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)

\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)

e: \(y=3^{x^2}\)

=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)

f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)

=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)

h: \(y=\left(x+1\right)\cdot e^{cosx}\)

=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)

=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)

\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)

a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)

\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)

\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)

b) \(y=\left(3x+1\right)^{\pi}\)

\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)

c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)

\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)

\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)

d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)

\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)

e) \(y=3^{x^2}\)

\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)

f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)

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

Các bài còn lại bạn tự làm nhé!

a: \(y'=4\cdot3x^2-3\cdot2x+2=12x^2-6x+2\)

b: \(y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}=\dfrac{x-1-x-1}{\left(x-1\right)^2}=\dfrac{-2}{\left(x-1\right)^2}\)

c: \(y'=-2\cdot\left(\sqrt{x}\cdot x\right)'\)

\(=-2\cdot\left(\dfrac{x+x}{2\sqrt{x}}\right)=-2\cdot\dfrac{2x}{2\sqrt{x}}=-2\sqrt{x}\)

d: \(y'=\left(3sinx+4cosx-tanx\right)\)'

\(=3cosx-4sinx+\dfrac{1}{cos^2x}\)

e: \(y'=\left(4^x+2e^x\right)'\)

\(=4^x\cdot ln4+2\cdot e^x\)

f: \(y'=\left(x\cdot lnx\right)'=lnx+1\)

a) Biểu thức \(f(x)\) có nghĩa khi và chỉ khi \(2x + 7 \ge 0,\)tức là khi \(x \ge \frac{{ - 7}}{2}.\)

Vậy tập xác định của hàm số này là \(D = \left[ { - \frac{7}{2}; + \infty )} \right.\)

b) Biểu thức \(f(x)\) có nghĩa khi và chỉ khi \({x^2} - 3x + 2 \ne 0,\)tức là khi \(x \ne 2,x \ne 1.\)

Vậy tập xác định của hàm số này là \(D = \mathbb{R}\backslash \left\{ {1;2} \right\}\)