Question

[Bùi Ánh Tuyết]

tính đạo hàm của hàm số sau

a) f(x) = 3x² – 5x;

b) f(x) = (1 + 2x)(x – 1);

gửi cô Hoàng Thi Hải lớp 11a4

Answer 1

[HT.Phong (9A5)]

`a)f(x)=3x^2-5x`

`f'(x)=(3x^2-5x)'=(3x^2)'-(5x)'`

`=6x-5`

`b)f(x)=(1+2x)(x-1)`

`=x-1+2x^2-2x`

`=2x^2-x-1`

`f'(x)=(2x^2-x-1)'=(2x^2)'-x'-1'`

`=4x-1`

Answer 2

[Lê Song Phương]

a) Xét \(x=a\in R\) bất kì. Ta có:

\(f^{\prime}\left(a\right)=\lim_{x\rarr a}\frac{f\left(x\right)-f\left(a\right)}{x-a}=\lim_{x\rarr a}\frac{3x^2-5x-3a^2+5a}{x-a}\)

\(=\lim_{x\rarr a}\frac{3\left(x-a\right)\left(x+a\right)-5\left(x-a\right)}{x-a}\)

\(=\lim_{x\rarr a}\left(3\left(x+a\right)-5\right)\)

\(=6a-5\)

Vậy \(f^{\prime}\left(x\right)=6x-5\)

b) Ta có \(f\left(x\right)=2x^2-x-1\)

Với \(x=b\in R\) bất kì, ta có:

\(f^{\prime}\left(b\right)=\lim_{x\rarr b}\frac{f\left(x\right)-f\left(b\right)}{x-b}=\lim_{x\rarr b}\frac{2x^2-x-1-2b^2+b+1}{x-b}\)

\(=\lim_{x\rarr b}\frac{2\left(x+b\right)\left(x-b\right)-\left(x-b\right)}{x-b}\)

\(=\lim_{x\rarr b}\left\lbrack2\left(x+b\right)-1\right\rbrack\)

\(=4b-1\)

Vậy \(f^{\prime}\left(x\right)=4x-1\)