Finding the derivative on the inverse of a function

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
Lebombo
Messages
144
Reaction score
0
I know the inverse of a function is found in two steps.

Isolate the independent variable and then switch the variables like this:

[tex][y = x^{3} +1] = [x = \sqrt[3]{y - 1}][/tex]

Then switch the variables to get: [tex]y = \sqrt[3]{x-1}[/tex]



However, when it comes to finding the derivative of the inverse of a function, is it true that the inverse of the function does not actually have to be found prior to differentiating?

For instance, I have this function: [tex]y=x^{5} + x + 1[/tex]

If all I do is switch the variables like so: [tex]x=y^{5}+y+1[/tex]

and then differentiate implicitly like so: [tex]\frac{d}{dx}[x] = \frac{d}{dx}[y^{5}+y+1][/tex]

[tex]= [1= (5y^4 +1)\frac{dy}{dx}] = [\frac{dy}{dx} = \frac{1}{5y^{4}+1}][/tex]



Two questions:

1) Is this the/a correct way to differentiate the inverse of a function?
2) If not correct, do I have to first find the inverse and then differentiate?
 
Physics news on Phys.org
Hi !

The inverse function of y(x) is x(y)
the derivative of y(x) with respect to x is dy/dx
the derivative of x(y) with respect to y is dx/dy = 1/(dy/dx)
Example :
y(x) = x^5 +x +1
dy/dx = 5 x^4+1
dx/dy = 1/(5 x^4 +1)