Hi Kate,

Since $f(3) = 4$ you know that $f^{-1}(4) = 3$ which is a constant so the derivative of $f^{-1}(4)$ is the derivative of $3$ which is zero, but I don't think this is what you were asked. I expect you were asked to

Find the derivative of $f^{-1}(x)$ at $x = 4$ if $f(3)=4$ and $f^{\;\prime}(3)=1/7.$

I am going to write $g(x) = f^{-1}(x)$ and since $f(3) = 4$ we know that $f^{-1}(4) = 3.$ Now apply the function $f$ to both sides of $g(x) = f^{-1}(x)$ to get

\[f\left(g(x)\right) = f\left(f^{-1}(x)\right) = x\]

and differentiate

\[f\left(g(x)\right) = x\]

using the chain rule to get

\[f^{\;\prime}\left(g(x)\right)\times g^{\prime}(x) = 1\]

Rewriting $g(x)$ as $f^{-1}(x)$ gives

\[f^{\;\prime}\left(f^{-1}(x)\right)\times \left( f^{-1}(x)\right)^{\prime}= 1\]

I hope this helps,
Penny