derivative of inverse functionDerivativeOfInverseFunction2007-05-10 10:19:312007-10-16 06:02:02Theoreminverse function% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}
\begin{thmplain}
If the real function $f$ has an inverse function $f_\leftarrow$ and the derivative of $f$ at the point \,
$x = f_\leftarrow(y)$\; is distinct from zero, then $f_\leftarrow$ is also differentiable at the point $y$ and
\begin{equation}
\label{d}
f_\leftarrow'(y) = \frac{1}{f'(x)}.
\end{equation}
That is, the derivatives of a function and its inverse function are inverse numbers of each other, provided that they have been taken at the points which correspond to each other.\\
\end{thmplain}
\begin{proof}
Now we have
$$f(f_\leftarrow(y)) = f(x) =y.$$
The derivatives of both sides must be equal:
$$\frac{d}{dy}\left[f(f_\leftarrow(y))\right] = \frac{d}{dy}y$$
Using the chain rule we get
$$f'(f_\leftarrow(y))\cdot f_\leftarrow'(y) = 1,$$
whence
$$f_\leftarrow'(y) = \frac{1}{f'(f_\leftarrow(y))}.$$
This is same as the asserted \PMlinkescapetext{equation} (\ref{d}).\\
\end{proof}
\textbf{Examples.}\; For simplicity, we express here the functions by symbols $y$ and the inverse functions by $x$.
\begin{enumerate}
\item $y = \tan{x}$,\; $x = \arctan{y}$;\;
$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{1+\tan^2{x}} =
\frac{1}{1+y^2}$
\item $y = \sin{x}$,\; $x = \arcsin{y}$;\;
$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{\cos{x}} =
\frac{1}{+\sqrt{1-\sin^2{x}}} = +\frac{1}{\sqrt{1-y^2}}$
\item $y = x^2$,\,\; $x = \pm\sqrt{y}$;\;
$\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{2x} =
\frac{1}{\pm2\sqrt{y}}$
\end{enumerate}
If the variable symbol $y$ in those results is changed to $x$, the results can be written
$$\frac{d}{dx}\arctan{x} = \frac{1}{1+x^2},\quad
\frac{d}{dx}\arcsin{x} = \frac{1}{\sqrt{1-x^2}},\quad
\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}.$$