Inverse functions and differentiation
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In mathematics, the inverse of a function y = f(x) is a function that, in some fashion, "undoes" the effect of f (see inverse function for a formal and detailed definition). The inverse of f is denoted f - 1. The statements y=f(x) and x=f-1(y) are equivalent.
Their two derivatives, assuming they exist, are reciprocal, as the Leibniz notation suggests; that is:
This is a direct consequence of the chain rule, since
and the derivative of x with respect to x is 1.
Writing explicitly the dependence of y on x and the point at which the differentiation takes place and using Lagrange's notation, the formula for the derivative of the inverse becomes
![\left[f^{-1}\right]'(a)=\frac{1}{f'\left[ f^{-1}(a) \right]}.](../../../math/f/1/b/f1b3905d469daa9001883076c0f00bfa.png)
Geometrically, a function and inverse function have graphs that are reflections, in the line y=x. This reflection operation turns the gradient of any line into its reciprocal.
Assuming that f has an inverse in a neighbourhood of x and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at x and have a derivative given by the above formula.
[edit] Examples
- y = x2 (for positive x) has inverse
.
At x=0, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.
- y = ex has inverse x = ln(y) (for positive y)
[edit] Additional properties
- Integrating this relationship gives
- This is only useful if the integral exists. In particular we need f'(x) to be non-zero across the range of integration.
- It follows that functions with continuous derivative have inverses in a neighbourhood of every point where the derivative is non-zero. This need not be true if the derivative is not continuous.
