



 
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
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,  


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