Confused about derivatives of inverse functions

[Scintillation]

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.

 

[HallsofIvy]

If y= f(x) is your original function, then it's inverse is y= f'(x). It looks to me like you are confusing the "y" written as a function of x and y written as the inverse function of x.

In any case, I can't understand what you are doing in your example. You have correctly found the derivative of [itex]y= x^3+ 2x- 8[/itex] with respect to x. Then you find the derivative of [itex]x= y^3+ 2y- 8[/itex] with respect to y. You shouldn't be surprized that you get exactly the same result- in the second case, x is exactly the same fuction of y that x is of y in the secondcase. If you were to solve the second equation for y, and differentiate that with respect to x, then you would have the derivative of the inverse. Of course, that would be an extremely complicated function- which is why the formula that tells us we can just invert the derivative of the functions is so useful.