%% Math 131, Fall, 2003
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\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Professor Hwang} \\
Derivatives of Inverse Functions
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We have seen how the chain rule can be used to calculate the
derivative of an inverse function. This sheet explains the general
situation and gives a geometric interpretation.

\paragraph{Symbolic Calculation} Suppose $f$~is differentiable, and
that~$f'$ is nowhere zero. Using the chain rule to differentiate the
identity $x=f\bigl(f^{-1}(x)\bigr)$ gives
\[
1=f'\bigl(f^{-1}(x)\bigr)\cdot(f^{-1})'(x),\qquad\text{or}\quad
(f^{-1})'(x)=\frac{1}{f'\bigl(f^{-1}(x)\bigr)}.
\]
Roughly, the derivative of the inverse function is the reciprocal of
the derivative of~$f$, but care is required with the points at which
the derivatives are evaluated.

\paragraph{Geometric Interpretation} The slope of a line $y=b+mx$
is~$m$, in that the rate of change of~$y$ with respect to~$x$ is~$m$.
If~$y$ is viewed as the independent variable, the slope becomes~$1/m$,
which is the rate of change of~$x$ with respect to~$y$. Fundamentally,
this is the reason why the derivatives of a function and its inverse
are reciprocals. In more detail, consider the following figures, which
depict the graphs of $f$~and $f^{-1}$:

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\noindent In the left-hand figure, the slope of the tangent line
is~$f'\bigl(f^{-1}(c)\bigr)$, the derivative of~$f$ at the indicated
point of the domain. In the right-hand figure, the slope of the
tangent line is $(f^{-1})'(c)$. Because the respective tangent lines
differ by reflection across the diagonal $y=x$, their slopes are
reciprocal. We find once again that
\[
(f^{-1})'(c) = \frac{1}{f'\bigl(f^{-1}(c)\bigr)}.
\]
Since the same formula holds for every~$c$ (as long as neither
function's derivative is zero), this formula is identical (except for
the name of the variable) to the formula found earlier.

\eject

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