In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function f in terms of the derivative of f. More precisely, if the inverse of f {\displaystyle f} is denoted as f − 1 {\displaystyle f^{-1}} , where f − 1 ( y ) = x {\displaystyle f^{-1}(y)=x} if and only if f ( x ) = y {\displaystyle f(x)=y} , then the inverse function rule is, in Lagrange's notation,
[ f − 1 ] ′ ( y ) = 1 f ′ ( f − 1 ( y ) ) . {\displaystyle \left[f^{-1}\right]'(y)={\frac {1}{f'\left(f^{-1}(y)\right)}}.}
This formula holds in general whenever f {\displaystyle f} is continuous and injective on an interval I, with f {\displaystyle f} being differentiable at f − 1 ( y ) {\displaystyle f^{-1}(y)} ( ∈ I {\displaystyle \in I} ) and where f ′ ( f − 1 ( y ) ) ≠ 0 {\displaystyle f'(f^{-1}(y))\neq 0} . The same formula is also equivalent to the expression
D [ f − 1 ] = 1 ( D f ) ∘ ( f − 1 ) , {\displaystyle {\mathcal {D}}\left[f^{-1}\right]={\frac {1}{({\mathcal {D}}f)\circ \left(f^{-1}\right)}},}
where D {\displaystyle {\mathcal {D}}} denotes the unary derivative operator (on the space of functions) and ∘ {\displaystyle \circ } denotes function composition. Geometrically, a function and inverse function have graphs that are reflections, in the line y = x {\displaystyle y=x} . This reflection operation turns the gradient of any line into its reciprocal. Assuming that f {\displaystyle f} has an inverse in a neighbourhood of x {\displaystyle x} and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at x {\displaystyle x} and have a derivative given by the above formula. The inverse function rule may also be expressed in Leibniz's notation. As that notation suggests,
d x d y d y d x = 1. {\displaystyle {\frac {dx}{dy}}\,{\frac {dy}{dx}}=1.}
This relation is obtained by differentiating the equation f − 1 ( y ) = x {\displaystyle f^{-1}(y)=x} in terms of x and applying the chain rule, yielding that:
d x d y d y d x = d x d x {\displaystyle {\frac {dx}{dy}}\,{\frac {dy}{dx}}={\frac {dx}{dx}}}
considering that the derivative of x with respect to x is 1.
Derivation Let f {\displaystyle f} be an invertible (bijective) function, let x {\displaystyle x} be in the domain of f {\displaystyle f} , and let y = f ( x ) . {\displaystyle y=f(x).} Let g = f − 1 . {\displaystyle g=f^{-1}.} So, f ( g ( y ) ) = y . {\displaystyle f(g(y))=y.} Differentiating this equation with respect to y {\displaystyle y} , and using the chain rule, one gets
f ′ ( g ( y ) ) ⋅ g ′ ( y ) = 1. {\displaystyle f'(g(y))\cdot g'(y)=1.}
That is,
g ′ ( y ) = 1 f ′ ( g ( y ) ) {\displaystyle g'(y)={\frac {1}{f'(g(y))}}}
or
[ f − 1 ] ′ ( y ) = 1 f ′ ( f − 1 ( y ) ) . {\displaystyle \left[f^{-1}\right]^{\prime }(y)={\frac {1}{f^{\prime }(f^{-1}(y))}}.}
Examples
y = x 2 {\displaystyle y=x^{2}} (for positive x) has inverse x = y {\displaystyle x={\sqrt {y}}} .
d y d x = 2 x ; d x d y = 1 2 y = 1 2 x {\displaystyle {\frac {dy}{dx}}=2x{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{2{\sqrt {y}}}}={\frac {1}{2x}}}
d y d x d x d y = 2 x ⋅ 1 2 x = 1. {\displaystyle {\frac {dy}{dx}}\,{\frac {dx}{dy}}=2x\cdot {\frac {1}{2x}}=1.}
At x = 0 {\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.
y = e x {\displaystyle y=e^{x}} (for real x) has inverse x = ln y {\displaystyle x=\ln {y}} (for positive y {\displaystyle y} )
d y d x = e x ; d x d y = 1 y = e − x {\displaystyle {\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}=e^{-x}}
d y d x d x d y = e x e − x = 1. {\displaystyle {\frac {dy}{dx}}\,{\frac {dx}{dy}}=e^{x}e^{-x}=1.}
Additional properties Integrating this relationship gives
f − 1 ( y ) = ∫ 1 f ′ ( f − 1 ( y ) ) d y + C . {\displaystyle {f^{-1}}(y)=\int {\frac {1}{f'({f^{-1}}(y))}}\,{dy}+C.}
This is only useful if the integral exists. In particular we need f ′ ( x ) {\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:
∫ f − 1 ( y ) d y = y f − 1 ( y ) − F ( f − 1 ( y ) ) + C {\displaystyle \int f^{-1}(y)\,{dy}=yf^{-1}(y)-F(f^{-1}(y))+C}
where F {\displaystyle F} denotes the antiderivative of f {\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 z = f ′ ( x ) {\displaystyle z=f'(x)} then we have, assuming f ″ ( x ) ≠ 0 {\displaystyle f''(x)\neq 0} : d d z [ f ′ ] − 1 ( z ) = 1 f ″ ( x ) {\displaystyle {\frac {d}{dz}}\left[f'\right]^{-1}(z)={\frac {1}{f''(x)}}} This can be shown using the previous notation y = f ( x ) {\displaystyle y=f(x)} . Then we have:
f ′ ( x ) = d y d x = d y d z d z d x = d y d z f ″ ( x ) ⇒ d y d z = f ′ ( x ) f ″ ( x ) {\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:
d d z [ f ′ ] − 1 ( z ) = d x d z = d y d z d x d y = f ′ ( x ) f ″ ( x ) 1 f ′ ( x ) = 1 f ″ ( x ) {\displaystyle {\frac {d}{dz}}[f']^{-1}(z)={\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 n ≥ 1 {\displaystyle n\geq 1} , with z = f ( n ) ( x ) {\displaystyle z=f^{(n)}(x)} , the nth derivative of f(x), and y = f ( n − 1 ) ( x ) {\displaystyle y=f^{(n-1)}(x)} , assuming f ( i ) ( x ) ≠ 0 for 0 < i ≤ n + 1 {\displaystyle f^{(i)}(x)\neq 0{\text{ for }}0<i\leq n+1} :
d d z [ f ( n ) ] − 1 ( z ) = 1 f ( n + 1 ) ( x ) {\displaystyle {\frac {d}{dz}}\left[f^{(n)}\right]^{-1}(z)={\frac {1}{f^{(n+1)}(x)}}}
Higher order derivatives The chain rule given above is obtained by differentiating the identity f − 1 ( y ) = x {\displaystyle f^{-1}(y)=x} with respect to y, where y = f ( x ) {\displaystyle y=f(x)} . One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x, one obtains
d 2 y d x 2 d x d y + d d x ( d x d y ) ( d y d x ) = 0 , {\displaystyle {\frac {d^{2}y}{dx^{2}}}\,{\frac {dx}{dy}}+{\frac {d}{dx}}\left({\frac {dx}{dy}}\right)\,\left({\frac {dy}{dx}}\right)=0,}
that is simplified further by the chain rule as
d 2 y d x 2 d x d y + d 2 x d y 2 ( d y d x ) 2 = 0. {\displaystyle {\frac {d^{2}y}{dx^{2}}}\,{\frac {dx}{dy}}+{\frac {d^{2}x}{dy^{2}}}\,\left({\frac {dy}{dx}}\right)^{2}=0.}
Replacing the first derivative, using the identity obtained earlier, we get
d 2 y d x 2 = − d 2 x d y 2 ( d y d x ) 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {d^{2}x}{dy^{2}}}\,\left({\frac {dy}{dx}}\right)^{3}}
which implies
d 2 x d y 2 = − d 2 y / d x 2 ( d y / d x ) 3 . {\displaystyle {\frac {d^{2}x}{dy^{2}}}=-{\frac {d^{2}y/dx^{2}}{\left(dy/dx\right)^{3}}}.}
Similarly for the third derivative we have
d 3 y d x 3 = − d 3 x d y 3 ( d y d x ) 4 − 3 d 2 x d y 2 d 2 y d x 2 ( d y d x ) 2 . {\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\left({\frac {dy}{dx}}\right)^{4}-3{\frac {d^{2}x}{dy^{2}}}\,{\frac {d^{2}y}{dx^{2}}}\,\left({\frac {dy}{dx}}\right)^{2}.}
Using the formula for the second derivative, we get
d 3 y d x 3 = − d 3 x d y 3 ( d y d x ) 4 + 3 ( d 2 y d x 2 ) 2 ( d y d x ) − 1 {\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\left({\frac {dy}{dx}}\right)^{4}+3\left({\frac {d^{2}y}{dx^{2}}}\right)^{2}\,\left({\frac {dy}{dx}}\right)^{-1}}
which implies
d 3 x d y 3 = − d 3 y / d x 3 ( d y / d x ) 4 + 3 ( d 2 y / d x 2 ) 2 ( d y / d x ) 5 . {\displaystyle {\frac {d^{3}x}{dy^{3}}}=-{\frac {d^{3}y/dx^{3}}{\left(dy/dx\right)^{4}}}+3{\frac {\left(d^{2}y/dx^{2}\right)^{2}}{\left(dy/dx\right)^{5}}}.}
In general, higher order derivatives of an inverse function can be expressed with Faà di Bruno's formula. Alternatively, the nth derivative can be written succinctly as an application of a differential operator:
[ f − 1 ] ( n ) ( y ) = [ ( 1 f ′ ( t ) d d t ) n t ] | t = f − 1 ( y ) . {\displaystyle \left[f^{-1}\right]^{(n)}(y)=\left.\left[\left({\frac {1}{f'(t)}}{\frac {d}{dt}}\right)^{n}t\right]\right|_{t=f^{-1}(y)}.}
In Lagrange's notation, the second and third derivatives from the examples above can be found as applying the operator and substituting t = f − 1 ( y ) {\displaystyle t=f^{-1}(y)} after:
[ f − 1 ] ″ ( y ) = [ 1 f ′ ( t ) d d t ( 1 f ′ ( t ) d d t t ) ] | t = f − 1 ( y ) = − f ″ ( f − 1 ( y ) ) [ f ′ ( f − 1 ( y ) ) ] 3 , {\displaystyle \left[f^{-1}\right]''(y)=\left.\left[{\frac {1}{f'(t)}}{\frac {d}{dt}}\left({\frac {1}{f'(t)}}{\frac {d}{dt}}t\right)\right]\right|_{t=f^{-1}(y)}=-{\frac {f''(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^{3}}},}
[ f − 1 ] ‴ ( y ) = [ 1 f ′ ( t ) d d t ( 1 f ′ ( t ) d d t ( 1 f ′