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'derivative of inverse function'
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| Title of object: |
derivative of inverse function |
| Canonical Name: |
DerivativeOfInverseFunction |
| Type: |
Topic |
| Created on: |
2007-05-10 10:19:31 |
| Modified on: |
2007-05-10 10:34:57 |
| Classification: |
msc:26A24 |
Revision comment (for changes between this and next version):
Preamble:
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\usepackage{amssymb}
\usepackage{amsmath}
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%\usepackage{psfrag}
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%\usepackage{graphicx}
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%\usepackage{xypic}
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\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
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Content:
If the real function $f$ has an inverse function $\overleftarrow{f}$ and the derivative of $f$ in the point \,
$x = \overleftarrow{f}(y)$\; in distinct from zero, then also $\overleftarrow{f}$ is differentiable in the point $y$ and
\begin{align}
\overleftarrow{f}'(y) = \frac{1}{f'(x)}.
\end{align}
That is, the derivatives of a function and its inverse function are inverse numbers of each other, provided that they have been taken in the points which correspond each other.\\
{\em Proof.}\, Now we have
$$f(\overleftarrow{f}(y)) = f(x) =y.$$
The derivatives of both sides must be equal:
$$\frac{d}{dy}\left[f(\overleftarrow{f}(y))\right] = \frac{d}{dy}y$$
Using the chain rule we get
$$f'(\overleftarrow{f}(y))\cdot\overleftarrow{f}'(y) = 1,$$
whence
$$\overleftarrow{f}'(y) = \frac{1}{f'(\overleftarrow{f}(y))}.$$
This is same as the asserted equation (1). |
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