Hi. I am reviewing for an exam, and this is a topic that I did not go through very thoroughly. 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.