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

b/ \(y'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\sqrt{x^3+1}-x^2+2\right)'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\dfrac{3x^2}{\sqrt{x^3+1}}-2x\right)\)c/ 

\(y'=14\left(x^6+2x-3\right)^6\left(x^6+2x-3\right)'=14\left(x^6+2x-3\right)^6\left(6x^5+2\right)\)

d/ \(y=\left(x^3-1\right)^{-\dfrac{5}{2}}\Rightarrow y'=-\dfrac{5}{2}\left(x^3-1\right)^{-\dfrac{7}{2}}\left(x^3-1\right)'=-\dfrac{15x^2}{2\sqrt{\left(x^3-1\right)^7}}\)

1. 

Cách 1: Tính bằng công thức

\(\left\{\begin{matrix} y=x-1(x>1)\\ y=1-x(x<1)\end{matrix}\right.\Rightarrow \left\{\begin{matrix} y'=1(x>1)\\ y'=-1(x<1)\end{matrix}\right.\)

Tóm gọn lại: $y'=\frac{|x-1|}{x-1}$

Cách 2: Tính bằng định nghĩa.

\(y'=\lim\limits_{x\to 1}\frac{|x-1|-0}{x-1}=\frac{|x-1|}{x-1}\)

2. Với $x\in (0;\pi)$ thì:

\(y=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{\cos x+1}{2}}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\cos ^2\frac{x}{2}}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{2}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\cos ^2\frac{x}{4}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\cos \frac{x}{4}}=\sqrt{\cos ^2\frac{x}{8}}=\cos \frac{x}{8}\)

\(\Rightarrow y'=-\frac{1}{8}\sin \frac{x}{8}\)

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:
\(x\in [-\sqrt{2}; \sqrt{2}]\Rightarrow x^2\leq 2\Rightarrow \sqrt{x^2+1}\leq \sqrt{3}\)

\(y=\frac{x+1}{\sqrt{x^2+1}}\geq \frac{x+1}{\sqrt{3}}\geq \frac{-\sqrt{2}+1}{\sqrt{3}}\)

Vậy $y_{\min}=\frac{-\sqrt{2}+1}{\sqrt{3}}$ khi $x=-\sqrt{2}$

$y^2=\frac{x^2+2x+1}{x^2+1}=1+\frac{2x}{x^2+1}$

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

$\Rightarrow y\leq \sqrt{2}$

Vậy $y_{\max}=\sqrt{2}$ khi $x=1$

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

\(b,y'=\left(\sqrt{x}+1\right)'\left(x^2+2\right)+\left(\sqrt{x}+1\right)\left(x^2+2\right)'\\ =\dfrac{x^2+2}{2\sqrt{x}}+\left(\sqrt{x}+1\right)\cdot2x\)

1) \(y=x^2-3\sqrt[]{x}+\dfrac{1}{x}\)

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

2) \(f\left(x\right)=\dfrac{x+9}{x+3}+4\sqrt[]{x}\)

\(\Rightarrow f'\left(x\right)=\dfrac{1.\left(x+3\right)-1\left(x+9\right)}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{x+3-x-9}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{-6}{\left(x+3\right)^2}+\dfrac{2}{\sqrt[]{x}}\)

\(\Rightarrow f'\left(1\right)=\dfrac{-6}{\left(1+3\right)^2}+\dfrac{2}{\sqrt[]{1}}=-\dfrac{3}{8}+2=\dfrac{13}{8}\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^2-1}+\sqrt[3]{\left(x-1\right)^3}}{\sqrt{x-1}}=\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^2-1\right)^{\dfrac{1}{2}}+x-1}{\left(x-1\right)^{\dfrac{1}{2}}}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^2-1\right)^{-\dfrac{1}{2}}.2+1}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}}\)

\(=\dfrac{1}{0}=+\infty\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt[3]{x}-1}{\sqrt{2}-\sqrt{x+1}}=\lim\limits_{x\rightarrow1^-}\dfrac{\left(x-1\right)\left(\sqrt{2}+\sqrt{x+1}\right)}{[\left(\sqrt[3]{x}\right)^2+\sqrt[3]{x}+1]\left(1-x\right)}=\lim\limits_{x\rightarrow1^-}\dfrac{-\left(\sqrt{2}+\sqrt{1+1}\right)}{1+1+1}=-\dfrac{2\sqrt{2}}{3}\)

\(f\left(1\right)=\sqrt{2}\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)\ne\lim\limits_{x\rightarrow1^+}f\left(x\right)\ne f\left(x\right)\)=> ham gian doan tai x=1

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

Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.

1.

\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)

\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)

2.

\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)

3.

\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)

\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)

4.

\(y'=-\dfrac{\left(\sqrt{sin^2\left(6-x\right)+4x}\right)'}{sin^2\left(6-x\right)+4x}=-\dfrac{\left[sin^2\left(6-x\right)+4x\right]'}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

5.

\(y'=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).\left[sin\left(\dfrac{2x-1}{4-x}\right)\right]'\)

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)

8.

\(y=tan^33x-\left(sin2x+cos3x\right)^5\)

\(\Rightarrow y'=3tan^23x.\left(tan3x\right)'-5\left(sin2x+cos3x\right)^4.\left(sin2x+cos3x\right)'\)

\(=\dfrac{9.tan^23x}{cos^23x}-5\left(sin2x+cos3x\right)^4.\left(2cos2x-3sin3x\right)\)

9.

\(y'=6cot^55x.\left(cot5x\right)'-4cos^33x.\left(cos3x\right)'+3cos3x\)

\(=-\dfrac{30.cot^55x}{sin^25x}+12cos^33x.sin3x+3cos3x\)