Câu hỏi của Lê Thành Nhân

Question

Nguyễn Lê Phước Thịnh · Accepted Answer

$f'\left(x\right)=\left(\dfrac{x^2+3x-5}{x+2}\right)'$

$=\dfrac{\left(x^2+3x-5\right)'\left(x+2\right)-\left(x^2+3x-5\right)\left(x+2\right)'}{\left(x+2\right)^2}$

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

$=\dfrac{2x^2+7x+6-x^2-3x+5}{\left(x+2\right)^2}=\dfrac{x^2+4x+11}{\left(x+2\right)^2}$

$f'\left(1\right)=\dfrac{1^2+4\cdot1+11}{\left(1+2\right)^2}=\dfrac{16}{9}$

Câu trả lời:

$f'\left(x\right)=\left(\dfrac{x^2+3x-5}{x+2}\right)'$

$=\dfrac{\left(x^2+3x-5\right)'\left(x+2\right)-\left(x^2+3x-5\right)\left(x+2\right)'}{\left(x+2\right)^2}$

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

$=\dfrac{2x^2+7x+6-x^2-3x+5}{\left(x+2\right)^2}=\dfrac{x^2+4x+11}{\left(x+2\right)^2}$

$f'\left(1\right)=\dfrac{1^2+4\cdot1+11}{\left(1+2\right)^2}=\dfrac{16}{9}$