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.

\(y=\left\{{}\begin{matrix}x-2\left(x\ge2\right)\\2-x\left(x\le2\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y'\left(2^+\right)=1\\y'\left(2^-\right)=-1\end{matrix}\right.\)

\(\Rightarrow y'\left(2^+\right)\ne y'\left(2^-\right)\Rightarrow\) không tồn tại đạo hàm tại \(x=2\)

b.

\(y=\left|x-2\right|^2=x^2-4x+4\Rightarrow y'=2x-4\)

\(\Rightarrow y'\left(2\right)=0\)

c.

\(y=\left\{{}\begin{matrix}4-x^2\left(\text{với }-2< x< 2\right)\\x^2-4\left(\text{với }x\ge2;x\le-2\right)\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}y'\left(2^+\right)=2x=4\\y'\left(2^-\right)=-2x=-4\end{matrix}\right.\)

\(\Rightarrow y'\left(2^+\right)\ne y'\left(2^-\right)\Rightarrow\) ko tồn tại đạo hàm tại \(x=2\)

d. Tương tự a và c

a) \(g'\left( x \right) = y' = {\left( {2x + \frac{\pi }{4}} \right)^,}.\cos \left( {2x + \frac{\pi }{4}} \right) = 2\cos \left( {2x + \frac{\pi }{4}} \right)\)

b) \(g'\left( x \right) =  - 2{\left( {2x + \frac{\pi }{4}} \right)^,}.\sin \left( {2x + \frac{\pi }{4}} \right) =  - 4\sin \left( {2x + \frac{\pi }{4}} \right)\)

a: \(y'=\left[tan\left(e^x+1\right)\right]'=\dfrac{\left(e^x+1\right)'}{cos^2\left(e^x+1\right)}=\dfrac{e^x}{cos^2\left(e^x+1\right)}\)

b: \(y'=\left(\sqrt{sin3x}\right)'\)

\(=\dfrac{\left(sin3x\right)'}{2\sqrt{sin3x}}=\dfrac{3\cdot cos3x}{2\sqrt{sin3x}}\)

c: \(y=cot\left(1-2^x\right)\)

=>\(y'=\left[cot\left(1-2^x\right)\right]'\)

\(=\dfrac{-2}{sin^2\left(1-2^x\right)}\cdot\left(-2^x\cdot ln2\right)\)

\(=\dfrac{2^{x+1}\cdot ln2}{sin^2\left(1-2^x\right)}\)

tham khảo:

a)y′=2\(^{3x-x^2}\).ln2.(3−2x)

b) y′\(\dfrac{4}{ln3}\).\(\dfrac{1}{4x+1}\).4=\(\dfrac{4}{\left(4x+1\right)ln3}\)

\(y=\dfrac{1}{3x^2-x-2}=\dfrac{1}{\left(x-1\right)\left(3x+2\right)}=\dfrac{1}{5}.\dfrac{1}{x-1}-\dfrac{3}{5}.\dfrac{1}{3x+2}\)

\(y'=\dfrac{1}{5}.\dfrac{\left(-1\right)^1.1!}{\left(x-1\right)^2}-\dfrac{3}{5}.\dfrac{\left(-1\right)^1.3^1.1!}{\left(3x+2\right)^2}\)

\(y''=\dfrac{1}{5}.\dfrac{\left(-1\right)^2.2!}{\left(x-1\right)^3}-\dfrac{3}{5}.\dfrac{\left(-1\right)^2.3^2.2!}{\left(3x+2\right)^3}\)

\(\Rightarrow y^{\left(n\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^n.n!}{\left(x-1\right)^{n+1}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^n.3^n.n!}{\left(3x+2\right)^{n+1}}\)

\(\Rightarrow y^{\left(2019\right)}=\dfrac{1}{5}.\dfrac{\left(-1\right)^{2019}.2019!}{\left(x-1\right)^{2020}}-\dfrac{3}{5}.\dfrac{\left(-1\right)^{2019}.3^{2019}.2019!}{\left(3x+2\right)^{2019}}\)

\(=\dfrac{2019!}{5}\left(\dfrac{3^{2020}}{\left(3x+2\right)^{2020}}-\dfrac{1}{\left(x-1\right)^{2020}}\right)\)

a) y' = 3.(x⁷- 5x²)².(x⁷- 5x²)' = 3.(x⁷ - 5x²)².(7x⁶ - 10x) = 3x.(x⁷ - 5x²)²(7x⁵ - 10).

b) y = 5x² - 3x⁴ + 5 - 3x² = -3x⁴ + 2x² + 5, do đó y' = -12x³ + 4x = -4x.(3x² - 1).

c) y' = = = .

d) y' = = = .

e) y' = 3. . = 3. = - ..

a.

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

b.

\(y'=\dfrac{\left(2sin^2x-1\right)'}{\left(2sin^2x-1\right).ln10}=\dfrac{2sin2x}{\left(2sin^2x-1\right)ln10}\)

c.

\(y'=\left(3x^2+3\right)3^{x^3+3x+1}.e^x.ln3+3^{x^3+3x+1}.e^x\)