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Finding the derivative on the inverse of a function

  1. Aug 23, 2013 #1
    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?
     
  2. jcsd
  3. Aug 24, 2013 #2
    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)
     
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