Derivatives of Inverse Functions

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 $f$ . The inverse of $f$ is denoted as $f^{-1}$ , where $f^{-1}(y)=x$ if and only if $f(x)=y$ .

Their two derivatives, assuming they exist, are reciprocal, as the Leibniz notation suggests; that is:

{\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}=1.

This relation is obtained by differentiating the equation $f^{-1}(y)=x$ in terms of $x$ and applying the chain rule, yielding that:

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

considering that 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, the formula for the derivative of the inverse becomes (in Lagrange's notation):

\left[f^{-1}\right]'(a)={\frac {1}{f'\left(f^{-1}(a)\right)}}

.

This formula holds in general whenever $f$ is continuous and injective on an interval $I$ , with $f$ being differentiable at $f^{-1}(a)$ ( $\in I$ ) and where $f'(f^{-1}(a))\neq 0$ . The same formula is also equivalent to the expression

{\mathcal {D}}\left[f^{-1}\right]={\frac {1}{({\mathcal {D}}f)\circ \left(f^{-1}\right)}},

where ${\mathcal {D}}$ denotes the unary derivative operator (on the space of functions) and $\circ$ denotes function composition.

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.