Question

Let m(x)=1/(f(x)^3) What is m'(-1)? f(-1)=-9 f'(-1)=7

Answer 1

\[m(x)=\frac{1}{f^3(x)}=f^{-3}(x)\]

Answer 2

\[m'(x)=-3f^{-4}(x)f'(x)\] by the chain rule

Answer 3

so
\[m'(1)=-3\times f^{-4}(1)\times f'(1)\]

Answer 4

now substitute the numbers to get your answer. it is
\[-3(-9)^{-4}\times 7\] or
\[\frac{-3\times 7}{(-9)^4}\]

Answer 5

in the second part why is it -3f^-4(x)f'(x) shouldnt it just be -3f^-4(x)

Answer 6

no
chain rule here. that is why you also need
\[f'(1)\]