Inverse functions and differentiation

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Rule:
{\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x))}

Example for arbitrary x_0 \approx 5.8:
{\color{CornflowerBlue}{f'}}(x_0) = \frac{1}{4}
{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x_0)) = 4~

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:

\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = 1.

This is a direct consequence of the chain rule, since

 \frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx}

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)}

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.

Examples[edit]

  • \,y = x^2 (for positive x) has inverse x = \sqrt{y}.
 \frac{dy}{dx} = 2x 
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{dx}{dy} = \frac{1}{2\sqrt{y}}=\frac{1}{2x}
 \frac{dy}{dx}\,\cdot\,\frac{dx}{dy}  =  2x \cdot\frac{1}{2x}  =  1.

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 = e^x (for real x) has inverse \,x = \ln{y} (for positive y)
 \frac{dy}{dx} = e^x
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{dx}{dy} = \frac{1}{y}
 \frac{dy}{dx}\,\cdot\,\frac{dx}{dy}  =  e^x \cdot \frac{1}{y}  =  \frac{e^x}{e^x}  =  1

Additional properties[edit]

  • Integrating this relationship gives
{f^{-1}}(x)=\int\frac{1}{f'({f^{-1}}(x))}\,{dx} + c.
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 a function that has a continuous derivative has an inverse in a neighbourhood of every point where the derivative is non-zero. This need not be true if the derivative is not continuous.

Higher derivatives[edit]

The chain rule given above is obtained by differentiating the identity x = f −1(f(x)) with respect to x. One can continue the same process for higher derivatives. Twice differentiating the identity with respect to x obtains,

 \frac{d^2y}{dx^2}\,\cdot\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2  =  0

or replacing the first derivative using the formula above,

 \frac{d^2y}{dx^2} = - \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^3.

Similarly for the third derivative:

 \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 -
3 \frac{d^2x}{dy^2}\,\cdot\,\frac{d^2y}{dx^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2

or using the formula for the second derivative,

 \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 +
3 \left(\frac{d^2x}{dy^2}\right)^2\,\cdot\,\left(\frac{dy}{dx}\right)^5

These formulas are generalized by the Faà di Bruno's formula.

These formulas can also be written using Lagrange's notation. If f and g are inverses, then

 g''(x) = \frac{-f''(g(x))}{[f'(g(x))]^3}

Example[edit]

  • \,y = e^x has the inverse \,x = \ln y. Using the formula for the second derivative of the inverse function,
 \frac{dy}{dx} = \frac{d^2y}{dx^2} = e^x = y 
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\left(\frac{dy}{dx}\right)^3 = y^3;

so that


\frac{d^2x}{dy^2}\,\cdot\,y^3 + y = 0
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{d^2x}{dy^2} = -\frac{1}{y^2}
,

which agrees with the direct calculation.

See also[edit]