# Homework Help: Derivatives of inverse functions-how two formulas relate?

1. May 3, 2013

### al_famky

Derivatives of inverse functions--how two formulas relate?

1. The problem statement, all variables and given/known data
I know two formulas for calculating the derivative of an inverse function, both of which I know how to derive, but I don't know how to relate them to one another.

2. Relevant equations
$\frac{df^{-1}(y)}{dy}$=$\frac{1}{\frac{df(x)}{dx}}$
$f'^{-1}(x)$=$\frac{1}{f'(f^{-1}(x))}$

3. The attempt at a solution
Well, for the first formula, it's because
$\frac{dy}{dx}$=$\frac{df(x)}{dx}$
$\frac{dx}{dy}$=$\frac{df^{-1}(y)}{y}$
so
you get the first equation above.

for the second formula, use
$\left(f(f^{-1}(x))\right)$=$x$
take the derivative of x of both sides
$\left(f'\left(f^{-1}(x)\right)\right)$$\left(f'^{-1}(x)\right)$=1
and with a simple algebraic manipulation you get the second formula

so, that being said...
I don't think that the y's and x's of the two formulas mean the same thing.
if i was given an equation like
$y=arctan(-x^{3}$)
then how would I use the two formulas (without using $arctan'x=\frac{1}{1+x^{2}}$)

I started out with the first formula
stating $x=arctan(-y^{3}$), so $-y^{3}=tanx$, $y=-tan^{\frac{1}{3}}x$
$\frac{df^{-1}(y)}{dy}$=$\frac{1}{\frac{dy}{dx}}$=$\frac{sec^{2}}{-\frac{1}{3}tan^{-\frac{2}{3}}x}$
or something like that... the answer is easy enough to get, but is it the answer i want? I calculated the derivative on the left with respect to y, and on the right i get an equation with x...which means i have to somehow change y to x, and there is no really convenient way (this formula was supposed to make things more convenient for calculation you know.)

for the second formula...things are easier, because both sides calculate derivates with respect to x, but that's where I get confused.
Are the two formulas related in any way? They should, but I can't wrap my mind around which y and which x they represent.

I don't know if i presented my problem clearly enough, but I hope someone would be willing to clarify the matter for me, thank you for your time.

Last edited: May 4, 2013