Question

tính f'(x) biết f(x) = \(\dfrac{x^2}{x+1}\)

tính y'(0) biết y = \(\dfrac{x}{x+1}\)

Answer 1

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

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

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

\(y'\left(0\right)=\dfrac{1}{\left(0+1\right)^2}=1\)