## Inverse Function Theorem

1. Introduction. So we have some function $y=f(x)$. The “Inverse Function” to $f$ is another function denoted $f^{-1}$ satisfying
(1)$f^{-1}(y)=x$.
WARNING: do not confuse $f^{-1}(y)$ with $f(y)^{-1}=1/f(y)$.

Observe we have $y=f(x)$ be a function of $x$. So what happens if we differentiate both sides of Equation (1) with respect to $x$? Lets try!

The right hand side is trivial
(2)$\displaystyle\frac{\mathrm{d}x}{\mathrm{d}x}=1$.
The left hand side is more interesting. We see, by the chain rule, it is
(3)$\displaystyle\frac{\mathrm{d}f^{-1}(y)}{\mathrm{d}x}=\frac{\mathrm{d}f^{-1}(y)}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}$.
But we do not know how to differentiate an inverse function…what to do?

Why not set Equations (2) and (3) equal? After all, the derivative of equal things turns out to be an equality! So we get
(4)$\displaystyle\frac{\mathrm{d}f^{-1}(y)}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}=1$.
If $\mathrm{d}y/\mathrm{d}x\not=0$, then we can divide both sides by it, obtaining
(5)$\displaystyle\frac{\mathrm{d}f^{-1}(y)}{\mathrm{d}y}=\frac{1}{\mathrm{d}y/\mathrm{d}x}$.
Thus we obtain the derivative of the inverse function without worrying about limits!

Inverse Function Theorem. Let $y=f(x)$ be a function and $f^{-1}(y)=x$ be its inverse function. Then
(5)$\displaystyle\frac{\mathrm{d}f^{-1}(y)}{\mathrm{d}y}=\frac{1}{\mathrm{d}y/\mathrm{d}x}$
provided $\mathrm{d}y/\mathrm{d}x\not=0$.

2. Example (Arcsine). Suppose we have want to find the derivative of $\arcsin(y)$, what do we do? First we use the inverse function theorem, writing $y=\sin(x)$ and thus
(6)$\displaystyle\frac{\mathrm{d}\arcsin(y)}{\mathrm{d}y}=\frac{1}{\mathrm{d}\sin(x)/\mathrm{d}x}$
The right hand side becomes
(7)$\displaystyle\frac{1}{\cos(x)}=\frac{1}{\sqrt{1-\sin^{2}(x)}}$
We plug in $y=\sin(x)$ obtaining
(7′)$\displaystyle\frac{1}{\cos(x)}=\frac{1}{\sqrt{1-\sin^{2}(x)}}=\frac{1}{\sqrt{1-y^{2}}}$.
Substitute this back into Equation (6) yields
(8)$\displaystyle\frac{\mathrm{d}\arcsin(y)}{\mathrm{d}y}=\frac{1}{\sqrt{1-y^{2}}}$
This is the derivative of $\arcsin(y)$.

Exercise 1. Recall $\arccos(y)$ is the inverse function for $y=\cos(x)$. What is $\displaystyle\frac{\mathrm{d}\arccos(y)}{\mathrm{d}y}$?

Exercise 2. Recall $\arctan(y)$ is the inverse function for $y=\tan(x)$. What is $\displaystyle\frac{\mathrm{d}\arctan(y)}{\mathrm{d}y}$?

Exercise 3. Suppose we have some function $f(x)=y$, and we use its tangent line $t(h)$ to a point $(x,y)$.
(a) What is the inverse function of $t(h)$?
(b) What is the tangent line to the inverse function at the same point?
(c) How does part (a) compare to part (b) of this exercise?

Exercise 4. Suppose $f^{-1}(y)$ is the inverse function to $f(x)=y$. What is
$\displaystyle\frac{\mathrm{d}^{2}f^{-1}(y)}{\mathrm{d}y^{2}}$
in terms of $f(x)$ and its derivatives?