8.2 Differentiating Inverse Functions

The first good news is that even though there is no general way to compute the value of the inverse to a function at a given argument, there is a simple formula for the derivative of the inverse of $f$ in terms of the derivative of $f$ itself.

In fact, the derivative of $f^{-1}$ is the reciprocal of the derivative of $f$, with argument and value reversed.

This is more or less obvious geometrically. The derivative of the function $y$ is $\frac{dy}{dx}$, while that of any inverse function to $y$ is $\frac{dx}{dy}$ which will be the reciprocal of the former at the value $x$ if evaluated at the $y$ value $y(x)$.

Let's prove this using algebra. All we have to do is to apply the chain rule to the defining property of $f^{-1}$ namely $f(f^{-1}(x)) = x$. By the chain rule we have $\frac{df^{-1}(x)}{dx}\frac{df(y)}{dy} = 1$ evaluated at $y = f^{-1}(x)$.

This means that the derivative of the inverse function is the reciprocal of the derivative of the function itself, evaluated at the value of the inverse function.

The argument seems simple enough but it is confusing. Can you use this rule to actually find derivatives of inverses without going nuts?

Let us see what this means for the exponential function and its inverse, $\ln(x)$. The derivative of the exponent function is itself, $\exp(x)$. Then the derivative of the logarithm function, $y = \ln(x)$ is the reciprocal of the exponent, evaluated at $\ln(x)$; this is $\frac{1}{\exp(\ln(x)}$, which is $\frac{1}{x}$. This latter claim follows from the definition of inverse which tells us that $\exp(\ln(x)) = x$.

Similarly, for the sine function, since its derivative at argument $x$ is $\cos(x)$, the derivative of $\arcsin(x)$ is the reciprocal of the cosine of itself, or $\frac{1}{\cos(\arcsin(x)}$.

You could leave it at that, but we generally reduce it to something slightly less ugly. A spreadsheet is as happy with this as with the result we end up with in the next paragraph. By the way, my spreadsheet gives the arcccosine function of argument A6 wherever I enter =acos(A6).

Since as we have seen in Chapter 7, $\cos x$ is $(1 - (\sin x)^2)^{\frac{1}{2}}$, and $\arcsin(\sin x)$ is $x$, we find that $\cos(\arcsin x)$ is $(1-x^2)^{\frac{1}{2}}$, and the reciprocal of that is the derivative of $\arcsin x$.

$(\arcsin x)' = \frac{1}{\cos(\arcsin x)} = \frac{1}{1-\sin(\arcsin x)^2)^{\frac{1}{2}}} = (1-x^2)^{-\frac{1}{2}}$

Similarly the derivative of $x^k$ is $kx^{k-1}$. This tells thus that the derivative of $x^{\frac{1}{k}}$ is the reciprocal of that evaluated at argument $x^{\frac{1}{k}}$. This is

$\frac{1}{k}(x^{\frac{1}{k}})^{1-k} = \frac{1}{k}x^{\frac{1}{k}-1}$

This is exactly the same result, $(x^a)' = ax^{a-1}$, that holds for integer powers,

In fact for any rational power, $a$, positive or negative, we have

$(x^a)' = ax^{a-1}$

We have already mentioned another piece of good news about inverse functions. Even though there is no obvious way to compute a particular value of one, at a particular argument, there is any easy way to compute the value of $f^{-1}(x)$ with a spreadsheet that you can actually perform in about a minute, once you know how, assuming you know how to compute $f$. All you have to do is reverse the order of the $x$ and $f(x)$ columns in doing an $xy$ scatter chart. By doing this you can see that the result gives a "multiple valued function" rather than an ordinary function, and can pick out your favorite single valued range for the inverse.

Exercises:

8.3 Use the fact proven above, $(x^{\frac{1}{k}})'= \frac{1}{k}x^{\frac{1}{k}-1}$ to find $(x^{\frac{j}{k}})'$. (you can use the multiple occurrence rule, or the product rule)

8.4 The tangent of an angle $z$, denoted as $\tan z$, is the ratio given by the sine divided by the cosine: $\tan x = \frac{\sin x}{\cos x}$. What is the derivative of $\tan z$? From it find the derivative of $\arctan z$ (called the arctangent of $z$), the inverse function to $\tan z$, (when the domain of $\tan z$ has been restricted to be from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.)