\(a,y'=8x^3-10x\\ \Rightarrow y''=24x^2-10\\ b,y'=e^x+xe^x\\ \Rightarrow y''=e^x+e^x+xe^x=2e^x+xe^x\)

\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)

Trước hết ta xét: \(g\left(x\right)=\dfrac{1}{x+a}=\left(x+a\right)^{-1}\) với a là hằng số bất kì

\(g'\left(x\right)=-1.\left(x+a\right)^{-2}=\left(-1\right)^1.1!.\left(x+a\right)^{-\left(1+1\right)}\) 

\(g''\left(x\right)=-1.\left(-2\right).\left(x+a\right)^{-3}=\left(-1\right)^2.2!.\left(x+a\right)^{-\left(2+1\right)}\)

Từ đó ta dễ dàng tổng quát được:

 \(g^{\left(n\right)}\left(x\right)=\left(-1\right)^n.n!.\left(x+a\right)^{-\left(n+1\right)}=\dfrac{\left(-1\right)^n.n!}{\left(x+a\right)^{n+1}}\)

Xét: \(f\left(x\right)=\dfrac{x^2+1}{x\left(x-2\right)\left(x+2\right)}=-\dfrac{1}{4}.\left(\dfrac{1}{x}\right)+\dfrac{5}{8}\left(\dfrac{1}{x+2}\right)+\dfrac{5}{8}\left(\dfrac{1}{x-2}\right)\)

Áp dụng công thức trên ta được:

\(f^{\left(30\right)}\left(1\right)=\dfrac{1}{4}.\dfrac{\left(-1\right)^{30}.30!}{1^{31}}+\dfrac{5}{8}.\dfrac{\left(-1\right)^{30}.30!}{\left(1+2\right)^{31}}+\dfrac{5}{8}.\dfrac{\left(-1\right)^{30}.30!}{\left(1-2\right)^{31}}\)

Bạn tự rút gọn kết quả nhé

\(f\left(x\right)=\dfrac{x^2+1}{x^3}-4x\) hay \(f\left(x\right)=\dfrac{x^2+1}{x^3-4x}\) bạn?

\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)

\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)

e,

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

f,

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

a: \(y'=\left(x^2-x\right)'=2x-1\)

\(y''=\left(2x-1\right)'=2\)

b: \(y'=\left(cosx\right)'=-sinx\)

\(y''=\left(-sinx\right)'=-cosx\)

Đáp án A

Ta có: y ' =    − 1 ( x − 3 ) 2 . ( x − 3 ) ' = − 1 ( x − 3 ) 2 y " = − 1 ( x − 3 ) 2 ' = − − 1 ( x − 3 ) 4 = 1 ( x − 3 ) 4 .2 ( ​ x − 3 ) = 2 ( x − 3 ) 3  ;  

⇒ y " ( 1 ) = 2 ( 1 − 3 ) 3 = − 1 4 .

a: \(y=x\cdot e^{2x}\)

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

\(=x\cdot\left(e^{2x}\right)'+x'\cdot\left(e^{2x}\right)\)

\(=e^{2x}+2\cdot x\cdot e^{2x}\)

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

\(=\left(e^{2x}\right)'+\left(2\cdot x\cdot e^{2x}\right)'\)

\(=4\cdot e^{2x}+4\cdot x\cdot e^{2x}\)

b: \(y=ln\left(2x+3\right)\)

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

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

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

a) Với \({x_0}\) bất kì, ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^3} - x_0^3}}{{x - {x_0}}}\\ = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {x - {x_0}} \right)\left( {{x^2} + x{x_0} + x_0^2} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \left( {{x^2} + x{x_0} + x_0^2} \right) = 3x_0^2\)

Vậy hàm số \(y = {x^3}\) có đạo hàm là hàm số \(y' = 3{x^2}\)

b) \(y' = \left( {{x^n}} \right)' = n{x^{n - 1}}\)