# Inverse functions and differentiation

In mathematics, the inverse of a function $y=f(x)$ is a function that, in some fashion, "undoes" the effect of $f$ (see inverse function for a formal and detailed definition). The inverse of $f$ is denoted as $f^{-1}$ , where $f^{-1}(y)=x$ if and only if $f(x)=y$ . Rule: f ′ ( x ) = 1 ( f − 1 ) ′ ( f ( x ) ) {\displaystyle {\color {CornflowerBlue}{f'}}(x)={\frac {1}{{\color {Salmon}{(f^{-1})'}}({\color {Blue}{f}}(x))}}} Example for arbitrary x 0 ≈ 5.8 {\displaystyle x_{0}\approx 5.8} : f ′ ( x 0 ) = 1 4 {\displaystyle {\color {CornflowerBlue}{f'}}(x_{0})={\frac {1}{4}}} ( f − 1 ) ′ ( f ( x 0 ) ) = 4   {\displaystyle {\color {Salmon}{(f^{-1})'}}({\color {Blue}{f}}(x_{0}))=4~}

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.