Đáp án: $f'(-2)=-\dfrac{19}{25}$
Giải thích các bước giải:
Ta có :
$f(x)=\dfrac{4x+5}{3x-1}$
$\to f'(x)=(\dfrac{4x+5}{3x-1})'$
$\to f'(x)=\dfrac{\left(4x+5\right)'\left(3x-1\right)-\left(3x-1\right)'\left(4x+5\right)}{\left(3x-1\right)^2}$
$\to f'(x)=\dfrac{4\left(3x-1\right)-3\left(4x+5\right)}{\left(3x-1\right)^2}$
$\to f'(x)=-\dfrac{19}{\left(3x-1\right)^2}$
$\to f'(-2)=-\dfrac{19}{\left(3\cdot2-1\right)^2}=-\dfrac{19}{25}$
