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Derivative of Inverse

  1. Nov 5, 2013 #1


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    Gold Member

    1. The problem statement, all variables and given/known data

    If f(x) = the third root of (x-8), find the derivative of its inverse.

    2. Relevant equations

    The derivative of its inverse = 1/f'(f^-1(x)) or 1 over its derivative at its inverse.

    3. The attempt at a solution

    I followed both the formula to verify my solution and also did some manual work in lieu of the formula. Basically, if g is the inverse of f, then f(g(x)) = x, and so the derivative of both sides is f'(g(x)g'(x) = 1, and since g'(x) is the derivative of the inverse of x, g'(x) = 1/f'(g(x)) and I get 3x^2. I have no idea if that's correct however. Note that g(x) = f^-1'(x); I'm typing g(x) since the latter notation gets messy really quickly when you don't bother to use LATEX. :p.

  2. jcsd
  3. Nov 5, 2013 #2


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    Staff: Mentor

    There is an easy way to check your work: calculate g(x), calculate the derivative, compare it with your result.
  4. Nov 5, 2013 #3
    Your final answer is correct but you actually don't need the inverse to find its derivative. There's a simpler way.

    We have
    where g(x) is the inverse of f(x).
    $$\Rightarrow g'(f(x))\cdot f'(x)=1 \Rightarrow g'(f(x))=\frac{1}{f'(x)}$$

    Substituting this
    $$g'(f(x))=3f^2(x) \Rightarrow g'(x)=3x^2$$
  5. Nov 6, 2013 #4


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    But in this particular case it is easier to find the inverse and then differentiate!
  6. Nov 6, 2013 #5
    Yes, it is. I just wanted to show an alternative approach because it won't be easy to find the inverse always. For example, I recently encountered the following problem:

    Let ##\displaystyle f(x)=\int_2^x \frac{dt}{\sqrt{1+t^4}}## and g(x) be the inverse of f(x). Find g'(0).

    Do you think it would be easy or even possible to find the inverse?
  7. Nov 6, 2013 #6
    Looks like [itex]\sqrt{17}[/itex] to me.
  8. Nov 19, 2013 #7
    Could you find a solution to this integral?
  9. Nov 19, 2013 #8
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