[[File:Umkehrregel 2.png|thumb|right|250px|The thick blue curve and the thick red curve are inverse to each other. A thin curve is the derivative of the same colored thick curve.
Inverse function rule:<br><math>{\color{CornflowerBlue}{f'(x) = \frac{1}({\color{Blue}{f(x))}</math><br><br>Example for arbitrary <math>x_0 \approx 5.8</math>:<br><math>{\color{CornflowerBlue}{f'(x_0) = \frac{1}{4}</math><br><math>{\color{Salmon}{\left[f^{-1}\right]'({\color{Blue}{f(x_0)) = 4~</math>]]
In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function in terms of the derivative of . More precisely, if the inverse of <math>f</math> is denoted as <math>f^{-1}</math>, where <math>f^{-1}(y) = x</math> if and only if <math>f(x) = y</math>, then the inverse function rule is, in Lagrange's notation,
:<math>\left[f^{-1}\right]'(y)=\frac{1}{f'\left( f^{-1}(y) \right)}.</math>
This formula holds in general whenever <math>f</math> is continuous and injective on an interval , with <math>f</math> being differentiable at <math>f^{-1}(y)</math>(<math>\in I</math>) and where<math>f'(f^{-1}(y)) \ne 0</math>. The same formula is also equivalent to the expression
:<math>\mathcal{D}\left[f^{-1}\right]=\frac{1}{(\mathcal{D} f)\circ \left(f^{-1}\right)},</math>
where <math>\mathcal{D}</math> denotes the unary derivative operator (on the space of functions) and <math>\circ</math> denotes function composition.
Geometrically, a function and inverse function have graphs that are reflections, in the line <math>y=x</math>. This reflection operation turns the gradient of any line into its reciprocal.
Assuming that <math>f</math> has an inverse in a neighbourhood of <math>x</math> and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at <math>x</math> 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,
:<math>\frac{dx}{dy}\,\frac{dy}{dx} = 1.</math>
This relation is obtained by differentiating the equation <math>f^{-1}(y)=x</math> in terms of and applying the chain rule, yielding that:
:<math>\frac{dx}{dy}\,\frac{dy}{dx} = \frac{dx}{dx}</math>
considering that the derivative of with respect to ' is 1.
Derivation
Let <math>f</math> be an invertible (bijective) function, let <math>x</math> be in the domain of <math>f</math>, and let <math>y=f(x).</math> Let <math>g=f^{-1}.</math> So, <math>f(g(y))=y.</math> Differentiating this equation with respect to , and using the chain rule, one gets
:<math>f'(g(y))\cdot g'(y)=1.</math>
That is,
:<math>g'(y)=\frac 1 {f'(g(y))}</math>
or
:<math>
\left[f^{-1}\right]^{\prime}(y) = \frac{1}{f^{\prime}(f^{-1}(y))}.
</math>
Examples
- <math>y = x^2</math> (for positive ) has inverse <math>x = \sqrt{y}</math>.
:<math> \frac{dy}{dx} = 2x
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{dx}{dy} = \frac{1}{2\sqrt{y=\frac{1}{2x} </math>
:<math> \frac{dy}{dx}\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x} = 1. </math>
At <math>x=0</math>, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.
- <math>y = e^x</math> (for real ) has inverse <math>x = \ln{y}</math> (for positive <math>y</math>)
:<math> \frac{dy}{dx} = e^x
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{dx}{dy} = \frac{1}{y} = e^{-x} </math>
:<math> \frac{dy}{dx}\,\frac{dx}{dy} = e^x e^{-x} = 1. </math>
Additional properties
- Integrating this relationship gives
::<math>{f^{-1(y)=\int\frac{1}{f'({f^{-1(y))}\,{dy} + C.</math>
:This is only useful if the integral exists. In particular we need <math>f'(x)</math> 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:
::<math> \int f^{-1}(y)\, {dy} = y f^{-1}(y) - F(f^{-1}(y)) + C </math>
:where <math> F </math> denotes the antiderivative of <math> f </math>.
- 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 <math> z = f'(x)</math> then we have, assuming <math> f(x) \neq 0</math>:<math display="block"> \frac{d}{dz}\left[f'\right]^{-1}(z) = \frac{1}{f(x)}</math>This can be shown using the previous notation <math> y = f(x)</math>. Then we have:
:<math display="block"> 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)}</math>Therefore:
:<math> \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)}</math>
By induction, we can generalize this result for any integer <math> n \ge 1</math>, with <math> z = f^{(n)}(x)</math>, the nth derivative of f(x), and <math> y = f^{(n-1)}(x)</math>, assuming <math> f^{(i)}(x) \neq 0 \text{ for } 0 < i \le n+1 </math>:
:<math> \frac{d}{dz}\left[f^{(n)}\right]^{-1}(z) = \frac{1}{f^{(n+1)}(x)}</math>
Higher order derivatives
The chain rule given above is obtained by differentiating the identity <math>f^{-1}(y)=x</math> with respect to , where <math>y=f(x)</math>. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to ', one obtains
:<math> \frac{d^2y}{dx^2}\,\frac{dx}{dy} + \frac{d}{dx} \left(\frac{dx}{dy}\right)\,\left(\frac{dy}{dx}\right) = 0, </math>
that is simplified further by the chain rule as
:<math> \frac{d^2y}{dx^2}\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\left(\frac{dy}{dx}\right)^2 = 0.</math>
Replacing the first derivative, using the identity obtained earlier, we get
:<math> \frac{d^2y}{dx^2} = - \frac{d^2x}{dy^2}\,\left(\frac{dy}{dx}\right)^3 </math>
which implies
:<math> \frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{\left(dy/dx\right)^3}. </math>
Similarly for the third derivative we have
:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\left(\frac{dy}{dx}\right)^4 -
3 \frac{d^2x}{dy^2}\,\frac{d^2y}{dx^2}\,\left(\frac{dy}{dx}\right)^2.</math>
Using the formula for the second derivative, we get
:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\left(\frac{dy}{dx}\right)^4 +
3 \left(\frac{d^2y}{dx^2}\right)^2\,\left(\frac{dy}{dx}\right)^{-1}</math>
which implies
:<math> \frac{d^3x}{dy^3} = - \frac{d^3y/dx^3}{\left(dy/dx\right)^4} + 3\frac{\left(d^2y/dx^2\right)^2}{\left(dy/dx\right)^5}. </math>
These formulas can also be written using Lagrange's notation:
:<math> \left[f^{-1}\right](y) = -\frac{f(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^3}, </math>
:<math> \left[f^{-1}\right](y) = -\frac{f(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^4} + 3\frac{\left[f(f^{-1}(y))\right]^2}{\left[f'(f^{-1}(y))\right]^5}. </math>
In general, higher order derivatives of an inverse function can be expressed with Faà di Bruno's formula. Alternatively, the th derivative can be written succinctly as:
:<math> \left[f^{-1}\right]^{(n)}(y) = \left[\left(\frac{1}{f'(t)}\frac{d}{dt}\right)^{n} t\right]_{t = f^{-1}(y)}. </math>
From this expression, one can also derive the th-integration of inverse function with base-point using Cauchy formula for repeated integration whenever <math>f(f^{-1}(y)) = y</math>:
:<math> \left[f^{-1}\right]^{(-n)}(y) = \frac{1}{n!} \left(f^{-1}(a)(y-a)^n + \int_{f^{-1}(a)}^{f^{-1}(y)}\left(y-f(u)\right)^{n}\,du\right). </math>
Example
- <math>y = e^x</math> has the inverse <math>x = \ln y</math>. Using the formula for the second derivative of the inverse function,
:<math> \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;</math>
so that
:<math>
\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},</math>
which agrees with the direct calculation.