Thread: derivatives of inverse functions

1. derivatives of inverse functions

this topic probably isn't supposed to be hard but i find it very confusing. my teacher went over it very briefly if not at all and my book doesn't help explain either. so it's my understanding that if you let y = f^-1 (x), then dy/dx = 1 / (dx/dy) and if you let y = f(x) and x = f^-1 (y), then dx/dy = 1 / (dy/dx). so derivatives of inverse functions can be written in 2 different ways? one of my problems asks to find the derivative of e^x (which i already know is e^x) by finding the derivative of the inverse of ln x. i did dy/dx = 1 / (1/y) = y. i also did dx/dy = 1 / (1/x) = x. how do i get e^x for either one? my equations are y = lnx and the inverse of that is x = lny.

2. Hmm, well you had $\displaystyle x=lny$ and that $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1}{y}} = y$

From the original function you can see that $\displaystyle x = lny$ is equivalent to $\displaystyle e^x = y$ so I think you may have done it correctly. However my prof barely touched upon this subject so I'm sort of unfamiliar with it as well.

3. Originally Posted by oblixps
this topic probably isn't supposed to be hard but i find it very confusing. my teacher went over it very briefly if not at all and my book doesn't help explain either. so it's my understanding that if you let y = f^-1 (x), then dy/dx = 1 / (dx/dy) and if you let y = f(x) and x = f^-1 (y), then dx/dy = 1 / (dy/dx). so derivatives of inverse functions can be written in 2 different ways? one of my problems asks to find the derivative of e^x (which i already know is e^x) by finding the derivative of the inverse of ln x. i did dy/dx = 1 / (1/y) = y. i also did dx/dy = 1 / (1/x) = x. how do i get e^x for either one? my equations are y = lnx and the inverse of that is x = lny.
You are getting your notation a bit confused. When talking about general functions, we might talk about f(x) and $\displaystyle f^{-1}(x)$ but when you have specific things like "$\displaystyle y= e^x$" it is best to write x= ln(y). From that, dx/dy= 1/y so [/tex]dy/dx= y= e^x[/tex]