Hi. I am reviewing for an exam, and this is a topic that I did not go through very thoroughly.(adsbygoogle = window.adsbygoogle || []).push({});

I understand how to calculate the derivative of an inverse function when I am given a point, as I simply use the equation (1/f((f^-1))'. So if I am given the equation y=x^3, for example, and am asked to find the derivative of the inverse at x=2, I simply do

1. 2=x^3, which helps me to find the corresponding x-value on the original function

2. find x=3√2, then put in 3√2 into dy/dx (3x^2).

3.The slope at the inverse would be the reciprocal of that.

I understand why that works.

However, when I am asked to find a general equation for the slope of the inverse, I get very confused. I don't understand why calculating dx/dy from the original function is different from calculating dy/dx on the inverse.

Here's an example (I know the picture is not of the best quality).

http://oi48.tinypic.com/25fh7k1.jpg

When I do dx/dy from the original function, I get 1/(3x^2 +2). But when I do dy/dx on the inverse function, I get 1/(3y^2 +2). I thought both of these should be the same. I am following Anton 9th edition, and I don't understand why they are different.

Since dy/dx= 1/(dx/dy), then it follows that dx=dx. So 1/(3x^2 +2)=(3y^2 +2). But that doesn't work either.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Confused about derivatives of inverse functions

**Physics Forums | Science Articles, Homework Help, Discussion**