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4. Derivatives of Inverse Functions

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Last week, we briefly showcased the technique of the logarithmic derivative, where we pulled a natural log out of a top hat to “pull apart” the base and power in a complicated exponent before differentiating. This idea of using implicit differentiation to make complicated derivatives slightly easier fits into a much broader set of problems, and our focus for this week will be applying it to inverse functions.

Consider the function $y = \sin ^{-1}(x)$. We may be interested in determining $\frac{dy}{dx}$ for one reason or another, but none of the derivative rules apply here. Note the chain rule does not apply — the $-1$ does not count as a power, and $\frac{dy}{dx}\neq -\frac{\cos x}{\sin^2 x}$.

If only we could get rid of that pesky $\sin ^{-1}$…which we can! Applying $\sin$ to both sides, we get for all $x$ in the domain of $\sin ^{-1}$ that $$\sin y = \sin \left( \sin ^{-1}(x) \right) = x.$$ Now differentiate the outermost dudes: $$\begin{align*} \frac{d}{dx}\sin y &= \frac{d}{dx} x \\ \cos y \cdot \frac{dy}{dx} &= 1 \\ \frac{dy}{dx} &= \frac{1}{\cos y} \\ &= \frac{1}{\cos \left( \sin ^{-1}(x) \right)}. \end{align*}$$

In this particular case, we can actually simplify this further in terms of $x$. Remember that $\sin ^{-1}(x)$ represents an angle on a triangle, and that $\sin \left( \sin ^{-1}(x) \right) = \frac{x}{1}$. Using soh cah toa, we may construct the following triangle:

Remember, $\sin ^{-1}(x)$ represents the angle for which sine (i.e. opposite over hypotenuse) is equal to $x = \frac{x}{1}$. Thus $x$ should be the opposite leg and $1$ should be the hypotenuse.

Now using this triangle, the adjacent leg is $\sqrt{1-x^2}$, and so $\cos \left( \sin ^{-1}(x) \right) = \sqrt{1-x^2}$. Putting this all together, we get $$\frac{dy}{dx} = \frac{1}{\cos \left( \sin ^{-1}(x) \right)}=\frac{1}{\sqrt{1-x^2}}.$$

You may notice a general pattern that can be applied to more than just inverse trig functions. Consider the following problem:

For the invertibility, we have $f’(x) = 1 + \frac{1}{x}$, which is always positive. $f$ is continuous and monotone, therefore it is invertible!

For the second part, it’s impossible for a mortal like me to solve for the inverse of $f$ explicitly, and I have no children to sacrifice. The main issue, of course, is that this elusive, mysterious, conniving, slimy, underhanded, lowlife of a function $f ^{-1}$ is preventing me from taking a derivative. But just like how we convrted a $\sin ^{-1}$ to a $\sin$ before, we can convert this $f ^{-1}$ into an $f$ as follows: $$f (y) = f \left( f ^{-1}(x) \right) = x.$$ Now we can take derivatives of the outermost guys: $$\frac{d}{dx}f(y) = f’(y) \cdot \frac{dy}{dx} = \left( 1 + \frac{1}{y} \right) \cdot \frac{dy}{dx} = 1.$$ Rearranging, we get that $$\frac{dy}{dx} = \frac{1}{1+\frac{1}{y}} = \frac{y}{y+1},$$ with the last equality from multiplying both sides of the fraction by $y$.

This is as far as we can go. We can rewrite $y = f ^{-1}(x)$, but that’s another dozen or so pen strokes my dainty fingers can’t afford. It is enough for this problem though — it only asks for the derivative at the point $(1, 1)$, i.e. at $x = 1$ and $y = 1$ (the value of $x$ is totally irrelevant). Thus, plugging in $y = 1$ yields $\frac{dy}{dx} = \frac{1}{2}$.

We can take this one step farther:

This statement is a bit gross, but it has a nice graphical interpretation. Recall that the graph of $f ^{-1}(x)$ is simply the reflection of $f(x)$ across the diagonal line $y=x$. It thus seems reasonable that the tangent lines to $f(x)$ also get reflected to tangent lines of $f ^{-1}(x)$.

The tangent line of $f ^{-1}(x)$ at the point $x$ is the reflection of the tangent line of $f(x)$ at the point $f ^{-1}(x)$. This reflection swaps “rise” and “run”, so the slopes of these tangent lines are reciprocals of one another! The formula in the inverse function theorem captures this geometric picture.