Find $g′\left(-\dfrac{1}{9}\right)$, where $g(x)$ is the inverse of $f(x)=\dfrac{x^{17}}{(x^2+8)}$.
What is $g\left(-\dfrac{1}{9}\right)$ and $g'\left(-\dfrac{1}{9}\right)$?
I tried setting $f(x)=-\dfrac{1}{9}$, but solving that would be far too time consuming. What is the correct way to go about this$?$
I am assuming that $f$ and $g$ are functions from $\mathbb{R}$ to $\mathbb{R}$. Now since they are inverses of each other we have
a) $f\circ g(x) = x$ and
b) $g\circ f(x) =x$.
Now we have that $f(-1) = -\dfrac{1}{9}$. So from (b) $g(f(-1))=-1$ that is $g\left(-\dfrac{1}{9}\right)=-1$
Also by chain rule of differentiation on (a) we get that $f'(g(x))g'(x)=1$ . So for $x=-\dfrac{1}{9}$ we get $f'(-1)g'\left(-\dfrac{1}{9}\right) =1$.
Can you complete it from here?