If $f(x)=x^3+e^\frac{x}{2}$, compute $\frac{d}{dx} \left(f^{-1}(x)\right) \text{at$~~x=1$}$ If $f(x)=x^3+e^\frac{x}{2}$
We need to compute
$$\frac{d}{dx} \left(f^{-1}(x)\right) \text{at$~~x=1$}$$
I tried to find $f^{-1}(x)$, but i couldn't
A hint that i got said
If $f(x)$ and $g(x)$ are inverse of each other then $$f^{'}(\alpha).g^{'}(\beta)=1$$
Where $(\alpha,\beta)$ is any point on $f(x)$
Now this did actually help me get the answer but how can we prove the above condition given as a Hint
All suggestions to deduce the condition are welcome
 A: Note that $f(0)=1$. (Also, $f$ is strictly increasing on $(0,\infty)$ so it is  one-to-one there).
Hence, $f^{-1}(1)=0$ and $(\frac d {dx} f^{-1}(x))_{x=1}=\frac  1 {f'(f^{-1}(1))}=\frac  1 {f'(0)}=2$.
A: I very well know that you can find the answer to your question (a derivative of the inverse function)
But I guess you are looking for the explanation for why $f'(\alpha)\times g'(\beta) = 1$ where $(\alpha, \beta) $ is point on $y = f(x)$
At first, let me tell you I'm bad at maths what I'll be explaining is just what I could imagine

*

*What is an inverse function:

If $y = f(x)$ is some function then the function obtained by changing the Y-axis with X-axis is inverse.

*

*Why $f'\times g' = 1$
The tangent to $y = f(x) $ must be perpendicular to tangent of $f^{-1}(x)$
Now, If the slope of the tangent to function$y = f(x)$ is $f'(x)_{\alpha}$ then the slope of the tangent to its inverse function must be equal to  reciprocal ($-\frac 1{f'(x)_{x = g(x)}}$)However, your view has been changed(rotated by 90-degrees) negative sign will not be there;

Thus,
$$g'(x) = \frac d{dx}f^{-1}(x) = \frac 1{f'(g(x))} = \frac 1{f'(f^{-1}(x))}$$
