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Inverse function rule

Formula for the derivative of an inverse function

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

considering that the derivative of x with respect to x is 1.

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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.} Derivating 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))}}}

( f − 1 ) ′ ( y ) = 1 f ′ ( f − 1 ( y ) ) . {\displaystyle (f^{-1})^{\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}}\,\cdot \,{\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}}\,\cdot \,{\frac {dx}{dy}}=e^{x}\cdot e^{-x}=1.}

Additional properties

Integrating this relationship gives

f − 1 ( x ) = ∫ 1 f ′ ( f − 1 ( x ) ) d x + C . {\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 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 ( x ) d x = x f − 1 ( x ) − F ( f − 1 ( x ) ) + C {\displaystyle \int f^{-1}(x)\,{dx}=xf^{-1}(x)-F(f^{-1}(x))+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 ( f ′ ) − 1 ( z ) d z = 1 f ″ ( x ) {\displaystyle {\frac {d(f')^{-1}(z)}{dz}}={\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 ( f ′ ) − 1 ( z ) d 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(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 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 ( f ( n ) ) − 1 ( z ) d z = 1 f ( n + 1 ) ( x ) {\displaystyle {\frac {d(f^{(n)})^{-1}(z)}{dz}}={\frac {1}{f^{(n+1)}(x)}}}

Higher derivatives

The chain rule given above is obtained by differentiating the identity f − 1 ( f ( x ) ) = x {\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

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}}}\,\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

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}}}\,\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

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}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{3}.}

Similarly for the third derivative:

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}}}\,\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,

d 3 y d x 3 = − d 3 x d y 3 ⋅ ( d y d x ) 4 + 3 ( d 2 x d y 2 ) 2 ⋅ ( d y d x ) 5 {\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

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

Example

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

d y d x = d 2 y d x 2 = e x = y ; ( d y d x ) 3 = y 3 ; {\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

d 2 x d y 2 ⋅ y 3 + y = 0 ; d 2 x d y 2 = − 1 y 2 {\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.

References

"Derivatives of Inverse Functions". oregonstate.edu. Archived from the original on 2021-04-10. Retrieved 2019-07-26. https://web.archive.org/web/20210410154136/https://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html ↩

Derivation

Examples

Additional properties

Higher derivatives

Example

See also

References