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3.3 Differentiating Inverse Functions

Syllabus
2020
Topic
3.3
Level
—

Reverse a Function, Reciprocate Its Rate

If f(b)=af(b)=a, then f−1(a)=bf^{-1}(a)=b: the inverse swaps the input and output. Differentiating the identity f(f−1(x))=xf(f^{-1}(x))=x with the chain rule shows that the inverse rate is the reciprocal of the original rate at the corresponding point.

\bigl(f^{-1}\bigr)'(a)=\frac{1}{f'\bigl(f^{-1}(a)\bigr)}=\frac{1}{f'(b)},\qquad f(b)=a\text{ and }f'(b)\ne 0

  1. To find (f−1)′(a)(f^{-1})'(a), locate bb such that f(b)=af(b)=a.
  2. Evaluate the original derivative f′(b)f'(b).
  3. Take its reciprocal: (f−1)′(a)=1/f′(b)(f^{-1})'(a)=1/f'(b).
    An explicit formula for f−1f^{-1} is unnecessary when the corresponding value bb is known.

Let f(x)=x3+xf(x)=x^3+x. To find (f−1)′(2)(f^{-1})'(2), note that f(1)=2f(1)=2, so f−1(2)=1f^{-1}(2)=1. Since f′(x)=3x2+1f'(x)=3x^2+1, (f−1)′(2)=1f′(1)=13(1)2+1=14.\bigl(f^{-1}\bigr)'(2)=\frac{1}{f'(1)}=\frac{1}{3(1)^2+1}=\frac14.

f−1(x)f^{-1}(x) means the inverse function, not the reciprocal 1/f(x)1/f(x). The reciprocal-rate formula requires an inverse on the relevant interval and f′(b)≠0f'(b)\ne 0; if the denominator is zero, this formula does not produce a finite inverse derivative.