4. Derivatives of Inverse Functions
≪ 3. Practise with Exponentials and Logarithms | Table of Contents | 5. Indeterminate Forms and L'Hôpital's Rule ≫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 \[\frac{dy}{dx} = \frac{1}{\cos \left( \sin ^{-1}(x) \right)}=\frac{1}{\sqrt{1-x^2}}.\]
Remark 1.
Whenever you’re told to take the derivative of an inverse trigonometric function, you can apply this technique. You will always end up with some standard trig functions with inverse trig functions inside them. Don’t freak out — draw the triangle as we’ve done above, label the angles and sides, and go from there.
You may notice a general pattern that can be applied to more than just inverse trig functions. Consider the following problem:
Example 2.
Let \(f(x) = x + \ln x\) for \(x > 0\).
- Show that \(f\) is invertible on its domain.
- If \(y = f ^{-1}(x)\), determine \(\frac{dy}{dx}\) at the point \((1, 1)\).
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:
Theorem 3. The Inverse Function Theorem
Let \(f(x)\) be a differentiable and invertible function. Then its inverse is differentiable whenever \(f’(x) \neq 0\), and \[\frac{d}{dx}f ^{-1}(x) = \frac{1}{f’\left( f ^{-1}(x) \right)}. \]
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.