Let $f (x) = x^{3}+x$ and $g (x) =x^{3} -x$ for all x.
I have to find derivative of $g\circ f^{-1}$ at $x=2$.
My textbook did this:
$(g \circ f^{-1})' (2) = \lim \limits_{h \to 0} \dfrac{g \circ f^{-1}(2 + h) - g \circ f^{-1}(2)}{h}$
$ = \lim \limits_{h \to 0} \dfrac{g(1 + \frac{h}{4}) - g(1)}{h}$
$= \lim \limits_{h \to 0} \dfrac{(1 + \frac{3h}{4}) - (1 + \frac{h}{4}) - 0}{h} $
$= \dfrac{1}{2}$
Can anyone please explain second step?
On
Using $f^{-1}\circ f=\operatorname{id}$ and applying the chainrule we find: $$g'\left(x\right)=\left(g\circ f^{-1}\circ f\right)'\left(x\right)=\left(g\circ f^{-1}\right)'\left(f\left(x\right)\right)f'\left(x\right)$$
Divide by $f'(x)$ to achieve: $$\left(g\circ f^{-1}\right)'\left(f\left(x\right)\right)=\frac{g'\left(x\right)}{f'\left(x\right)}$$
Substitute $f^{-1}(x)$ for $x$ to achieve on base of $f\circ f^{-1}=\operatorname{id}$: $$\left(g\circ f^{-1}\right)'\left(x\right)=\frac{g'\left(f^{-1}\left(x\right)\right)}{f'\left(f^{-1}\left(x\right)\right)}$$
$f'(2)=4$, so $\bigl(f^{-1}\bigr)'\bigl(f^{-1}(2)\bigr)=\dfrac{1}{4}$. Btw, there's a typo at the beginning of the formula: it's $\bigl(g\circ f^{-1}\bigr)'(2)$ that is being computed.
More generally, you can prove that $$\bigl(g\circ f^{-1}\bigr)'(x)=\dfrac{g'\bigl(f^{-1}(x)\bigr)}{f'\bigl(f^{-1}(x)\bigr) }. $$
Indeed, the chain rule results in$$\Bigl(g\circ f^{-1}\Bigr)'(x)= g'\bigl(f^{-1}(x)\bigr)\cdot\bigl(f^{-1}\bigr)'(x)= g'\bigl(f^{-1}(x)\bigr)\cdot\frac{1}{f'\bigl(f^{-1}(x)\bigr)} $$ since $\bigl( f^{-1}\bigr)'(x)= \dfrac{1}{f'(y)}$, where $f(y)=x$.