Inverse function rule

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

• $\,y=x^{2}$ (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

• 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

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. Differentiating the identity twice with respect to x , one obtains

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

that is simplified further by the chain rule as

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

Replacing the first derivative, using the identity obtained earlier, we get

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

Similarly for the third derivative:

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

or using the formula for the second derivative,

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

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

• $\,y=e^{x}$ has the inverse $\,x=\ln y$. Using the formula for the second derivative of the inverse function,

${\frac {dy}{dx}}={\frac {d^{2}y}{dx^{2}}}=e^{x}=y{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}\left({\frac {dy}{dx}}\right)^{3}=y^{3};$

so that

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

which agrees with the direct calculation.

See also

• Calculus
• Inverse functions
• Chain rule
• Inverse function theorem
• Implicit function theorem
• Integration of inverse functions