Question 10.12

Solutions

leumasicOfficial

3 months ago

We have

\begin{align*} \left(\frac{1}{g}\right)'(x)&=(f\circ g)'(x) \\ &=f'(g(x))\cdot g'(x) \\ &=\frac{-1}{g(x)^2}\cdot g'(x). \end{align*}

0

Submit a solution

Optional • Markdown

Sign in to share your solution for this question.

Navigate

Q 10.12

Navigate

Q 10.12