Let ${\displaystyle f}$  be an invertible (bijective) function, let ${\displaystyle x}$  be in the domain of ${\displaystyle f}$ , and let ${\displaystyle y}$  be in the codomain of ${\displaystyle f}$ . Since f is a bijective function, ${\displaystyle y}$  is in the range of ${\displaystyle f}$ . This also means that ${\displaystyle y}$  is in the domain of ${\displaystyle f^{-1}}$ , and that ${\displaystyle x}$  is in the codomain of ${\displaystyle f^{-1}}$ . Since ${\displaystyle f}$  is an invertible function, we know that ${\displaystyle f(f^{-1}(y))=y}$ . The inverse function rule can be obtained by taking the derivative of this equation.

${\displaystyle {\dfrac {\mathrm {d} }{\mathrm {d} y}}f(f^{-1}(y))={\dfrac {\mathrm {d} }{\mathrm {d} y}}y}$ 

The right side is equal to 1 and the chain rule can be applied to the left side:

${\displaystyle {\begin{aligned}{\dfrac {\mathrm {d} \left(f(f^{-1}(y))\right)}{\mathrm {d} \left(f^{-1}(y)\right)}}{\dfrac {\mathrm {d} \left(f^{-1}(y)\right)}{\mathrm {d} y}}&=1\\{\dfrac {\mathrm {d} f(f^{-1}(y))}{\mathrm {d} f^{-1}(y)}}{\dfrac {\mathrm {d} f^{-1}(y)}{\mathrm {d} y}}&=1\\f^{\prime }(f^{-1}(y))(f^{-1})^{\prime }(y)&=1\end{aligned}}}$ 

Rearranging then gives

${\displaystyle (f^{-1})^{\prime }(y)={\frac {1}{f^{\prime }(f^{-1}(y))}}}$ 

Rather than using ${\displaystyle y}$  as the variable, we can rewrite this equation using ${\displaystyle a}$  as the input for ${\displaystyle f^{-1}}$ , and we get the following:^[2]

${\displaystyle (f^{-1})^{\prime }(a)={\frac {1}{f^{\prime }\left(f^{-1}(a)\right)}}}$ 

• ${\displaystyle y=x^{2}}$  (for positive x) has inverse ${\displaystyle x={\sqrt {y}}}$ .

${\displaystyle {\frac {dy}{dx}}=2x{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{2{\sqrt {y}}}}={\frac {1}{2x}}}$ 

${\displaystyle {\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=2x\cdot {\frac {1}{2x}}=1.}$ 

At ${\displaystyle x=0}$ , however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.

• ${\displaystyle y=e^{x}}$  (for real x) has inverse ${\displaystyle x=\ln {y}}$  (for positive ${\displaystyle y}$ )

${\displaystyle {\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}=e^{-x}}$ 

${\displaystyle {\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=e^{x}\cdot e^{-x}=1.}$ 

• Integrating this relationship gives

${\displaystyle {f^{-1}}(x)=\int {\frac {1}{f'({f^{-1}}(x))}}\,{dx}+C.}$ 

This is only useful if the integral exists. In particular we need ${\displaystyle f'(x)}$  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.

• Another very interesting and useful property is the following:

${\displaystyle \int f^{-1}(x)\,{dx}=xf^{-1}(x)-F(f^{-1}(x))+C}$ 

Where ${\displaystyle F}$  denotes the antiderivative of ${\displaystyle f}$ .

• The inverse of the derivative of f(x) is also of interest, as it is used in showing the convexity of the Legendre transform.

Let ${\displaystyle z=f'(x)}$  then we have, assuming ${\displaystyle f''(x)\neq 0}$ :$${\displaystyle {\frac {d(f')^{-1}(z)}{dz}}={\frac {1}{f''(x)}}}$$ This can be shown using the previous notation ${\displaystyle y=f(x)}$ . Then we have:

$${\displaystyle f'(x)={\frac {dy}{dx}}={\frac {dy}{dz}}{\frac {dz}{dx}}={\frac {dy}{dz}}f''(x)\Rightarrow {\frac {dy}{dz}}={\frac {f'(x)}{f''(x)}}}$$ Therefore:

${\displaystyle {\frac {d(f')^{-1}(z)}{dz}}={\frac {dx}{dz}}={\frac {dy}{dz}}{\frac {dx}{dy}}={\frac {f'(x)}{f''(x)}}{\frac {1}{f'(x)}}={\frac {1}{f''(x)}}}$ 

By induction, we can generalize this result for any integer ${\displaystyle n\geq 1}$ , with ${\displaystyle z=f^{(n)}(x)}$ , the nth derivative of f(x), and ${\displaystyle y=f^{(n-1)}(x)}$ , assuming ${\displaystyle f^{(i)}(x)\neq 0{\text{ for }}0<i\leq n+1}$ :

${\displaystyle {\frac {d(f^{(n)})^{-1}(z)}{dz}}={\frac {1}{f^{(n+1)}(x)}}}$ 

The chain rule given above is obtained by differentiating the identity ${\displaystyle f^{-1}(f(x))=x}$  with respect to x. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x, one obtains

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

that is simplified further by the chain rule as

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

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

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

Similarly for the third derivative:

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

or using the formula for the second derivative,

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

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

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

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

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

so that

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

which agrees with the direct calculation.

• Calculus – Branch of mathematics
• Chain rule – For derivatives of composed functions
• Differentiation of trigonometric functions – Mathematical process of finding the derivative of a trigonometric function
• Differentiation rules – Rules for computing derivatives of functions
• Implicit function theorem – On converting relations to functions of several real variables
• Integration of inverse functions – Mathematical theorem, used in calculusPages displaying short descriptions of redirect targets
• Inverse function – Mathematical concept
• Inverse function theorem – Theorem in mathematics
• Table of derivatives – Rules for computing derivatives of functionsPages displaying short descriptions of redirect targets
• Vector calculus identities – Mathematical identities

• Marsden, Jerrold E.; Weinstein, Alan (1981). "Chapter 8: Inverse Functions and the Chain Rule". Calculus unlimited (PDF). Menlo Park, Calif.: Benjamin/Cummings Pub. Co. ISBN 0-8053-6932-5.