a.

\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)

b.

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

c.

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

d.

\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)

e.

\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)

g.

\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)

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!

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

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

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

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

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: \(y=ln^2x\)

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

1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)

2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)

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

a: ĐKXĐ: \(\left(x+2\right)\left(x+3\right)>=0\)

=>\(\left[{}\begin{matrix}x>=-2\\x< =-3\end{matrix}\right.\)

\(y=\sqrt{\left(x+2\right)\left(x+3\right)}=\sqrt{x^2+5x+6}\)

=>\(y'=\dfrac{\left(x^2+5x+6\right)'}{2\sqrt{x^2+5x+6}}=\dfrac{2x+5}{2\sqrt{x^2+5x+6}}\)

y'>0

=>\(\dfrac{2x+5}{2\sqrt{x^2+5x+6}}>0\)

=>2x+5>0

=>\(x>-\dfrac{5}{2}\)

Kết hợp ĐKXĐ, ta được: x>=-2

Đặt y'<0

=>2x+5<0

=>2x<-5

=>\(x< -\dfrac{5}{2}\)

Kết hợp ĐKXĐ, ta được: x<=-3

Vậy: Hàm số đồng biến trên \([-2;+\infty)\) và nghịch biến trên \((-\infty;-3]\)

b: ĐKXĐ: \(\dfrac{2x+1}{x-3}>=0\)

=>\(\left[{}\begin{matrix}x>3\\x< =-\dfrac{1}{2}\end{matrix}\right.\)

\(y=\sqrt{\dfrac{2x+1}{x-3}}\)

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

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

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

\(=-\dfrac{\dfrac{7}{\left(x-3\right)^2}}{2\sqrt{\dfrac{2x+1}{x-3}}}< 0\forall x\) thỏa mãn ĐKXĐ, trừ x=-1/2 ra

=>Hàm số luôn đồng biến trên \(\left(3;+\infty\right);\left(-\infty;-\dfrac{1}{2}\right)\)

c:

ĐKXĐ: x>=-3

 \(y=\left(x+1\right)\sqrt{x+3}\)

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

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

=>\(y'=\sqrt{x+3}+\dfrac{x+1}{2\sqrt{x+3}}\)

=>\(y'=\dfrac{2x+6+x+1}{2\sqrt{x+3}}=\dfrac{3x+7}{2\sqrt{x+3}}\)

Đặt y'>0

=>3x+7>0

=>x>-7/3

Kết hợp ĐKXĐ, ta được: x>-7/3

Đặt y'<0

3x+7<0

=>x<-7/3

Kết hợp ĐKXĐ, ta được: \(-3< x< -\dfrac{7}{3}\)

Vậy: Hàm số đồng biến trên \(\left(-\dfrac{7}{3};+\infty\right)\) và nghịch biến trên \(\left(-3;-\dfrac{7}{3}\right)\)

d: \(y=\dfrac{x-1}{x^2+1}\)(ĐKXĐ: \(x\in R\))

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

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

Đặt y'>0

=>\(-x^2+2x+1>0\)

=>\(1-\sqrt{2}< x< 1+\sqrt{2}\)

Đặt y'<0

=>\(-x^2+2x-1< 0\)

=>\(\left[{}\begin{matrix}x>1+\sqrt{2}\\x< 1-\sqrt{2}\end{matrix}\right.\)

Vậy: hàm số đồng biến trên khoảng \(\left(1-\sqrt{2};1+\sqrt{2}\right)\)

hàm số nghịch biến trên khoảng \(\left(1+\sqrt{2};+\infty\right);\left(-\infty;1-\sqrt{2}\right)\)

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.