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Relationship between the derivative of f and the derivative of f inverse

  1. Feb 4, 2010 #1
    Since [tex]f^{-1}[/tex] is a reflection of [tex]f[/tex] about the line y=x, we can say that the derivative (slope of tangent at any point) of [tex]f^{-1}[/tex] is the reciprocal of the derivative of [tex]f[/tex].

    Now suppose [tex]y=x^3+2x^2+4[/tex]

    so that [tex]\frac{dy}{dx} = 3x^2+4x[/tex]

    But then why am I wrong in saying that [tex] ( f^{-1} )'(x) = \frac{1}{ \frac{dy}{dx} } = \frac{1}{3x^2+4x}?[/tex]
  2. jcsd
  3. Feb 4, 2010 #2
    Because the inverse's derivative is a function of y, not x.
  4. Feb 5, 2010 #3
    But if we look at the curve of the inverse, we'll see that to find the tangent at any point on the curve, we'll have to replace x with with that point. I mean, the tangent is still dy/dx, because the axes are the same: y and x.
  5. Feb 5, 2010 #4
    The tangent line will be the same, but its slope will change, because now you calculate the rate of change of x wrt y.

    When you wrote your formula, you treated both f^-1 and y as a function of the same variable x (or to be more accurate, functions of the same range), while it's not true.

    To deliver this notion via the formula it's more accurate to write:

    [tex]\frac{d}{dy}f^{-1}(y)|_{y=y_{0}}=(\frac{d}{dx}f(x))^{-1}|_{x=f^{-1}(y_{0})} [/tex]
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