Let $f(x)=3+x^2+\tan\frac{\pi x}2$, where $-1I am tasked with the following problem:

Let $f(x)=3+x^2+\tan\frac{\pi x}2$, where $-1<x<1$. Find $(f^{-1})'(3)$.

However, when I attempt to find the inverse of the function, it seems that none can be formulated. I worked out that the derivative is $\frac{\pi \sec ^2\left(\frac{\pi x}{2}\right)}{2}+2x$, but I find that both of these functions seem to head to infinity.
 A: You do not need to find the inverse expression. All you need to note is that $f(0)=3$, so
$$(f^{-1})'(3)=1/f'(0)=\frac1{\pi/2+2\cdot0}=\frac2\pi$$
A: Say, $$f^{-1}(x)=g(x)$$
$$\Rightarrow g(f(x))=x$$
$$\Rightarrow  g'(f(x))f'(x)=1$$
at $x=0$
$$f(0)=3 $$ and $$f'(0)=\frac{\pi}{2}$$
now,$$  g'(f(0))f'(0)=1$$
$$\Rightarrow  g'(3)\frac{\pi}{2}=1$$
$$\Rightarrow  g'(3)=\frac{2}{\pi}$$
$$\Rightarrow  (f^{-1})'(3)=\frac{2}{\pi}$$
A: If $f(x) =y $ then $$(f^{-1} )' (y) =\frac{1}{f'(x)}$$
Here we have $f(x) =3 $ and $x\in (-1,1) $ if and only if $x=0.$
$$f'(x) =2x +\frac{1}{\cos^2 \frac{\pi x}{2}} \cdot \frac{\pi}{2}$$
Hence $$(f^{-1} )' (3) =\frac{1}{f'(0)}=\frac{2}{\pi} $$
A: You definitely don't need to find the general form of $f^{-1}(x)$. Just denote $g=f^{-1}$ and take teh derivative of
$$g(f(x))=x$$
which is
$$g'(f(x))f'(x)=1$$
Just note that $f(0)=3$ so one has
$$g'(3)f'(0)=1$$
hence
$$\left(f^{-1}\right)'(3)={1\over f'(0)}={2\over\pi}$$
A: $$f(x)=3+x^2+\tan(\frac{\pi x}{2})$$
Let $g(x)=f^{-1}(x)$
$$g(f(x))=x$$
$$\implies g'(f(x))f'(x)=1$$
$$\implies g'(f(x))=\frac{1}{2x+\frac{\pi}{2}\sec^2(\frac{\pi x}{2})}$$

$$f(x)=3 \implies x=0$$

$$\implies g'(3)=\frac{1}{\frac{\pi}{2}\sec^2(0)}=\boxed{\frac{2}{\pi}}$$
A: We have $( x=0,f(0)= y=3)$ before reflecting about $x=y$
and slope evaluated at $x=0$ is $2x+ \sec \dfrac{\pi x}{2} = \dfrac{\pi}{2} $ and since
$$ x\rightarrow y,\; \dfrac{dy}{dx}\rightarrow \dfrac{dx}{dy}$$
the slope of inverse function at $(x=3)$ is its reciprocal, i.e., $\dfrac{2}{\pi}.$
