Section 3.7 Derivatives of Inverse Functions
Reference.
The Chain Rule for compositions, Equation (3.6.2) in Section 3.6, gives us a rule for the derivative of the inverse a function, because a function and its inverse are connected by composition: For \(y = g(x) = f^{-1}(x)\) the inverse of \(x = f(y)\text{,}\) their composition brings you back where you started:
\begin{equation*}
f(g(x)) = x
\end{equation*}
The Chain Rule then connects their derivatives in a way that looks quite simple in Leibniz notation:
- the derivative of \(y=g(x)\) that we seek is \(g'(x) = dy/dx\text{,}\)
- the derivative of its inverse \(x=f(y)\) is \(f'(y) = dx/dy\text{,}\) so
- the derivative of their composition \(x=f(g(x))\) is\begin{equation*} 1 = \frac{dx}{dx} = \frac{dx}{dy} \frac{dy}{dx}\text{:} \end{equation*}
- thus in words, the derivative of the inverse is the reciprocal of the derivative:\begin{equation} \frac{dy}{dx} = \frac{1}{dx/dy}\text{.}\tag{3.7.1} \end{equation}
This is just as you might guess from thinking of derivative notations like \(dy/dx\) as being arithmetic with small quantities \(dx\) and \(dy\text{!}\)
However, we need to be careful with the fact that the function and its inverse have different input arguments; \(x\) and \(y = f(x)\) respectively. The Lagrange "\(f'\)" notation makes the issue clearer:
\begin{equation*}
g'(x) = \frac{1}{f'(y)}
\end{equation*}
showing that this gives the derivative — a function of \(x\) — in terms of \(y\) instead. To get a formula in terms of \(x\text{,}\) eliminate \(y\) by using \(y = g(x)\text{:}\)
\begin{equation*}
g'(x) = \frac{1}{f'(g(x))}
\end{equation*}
Finally, to put it all in terms of the original function \(f\) whose derivative we already know,
\begin{equation}
[f^{-1}]'(x) = \frac{1}{f'(f^{-1}(x))}\text{.}\tag{3.7.2}
\end{equation}
Let's start with an example where we can check the answer: the derivative of the square root function \(y = g(x) = \sqrt{x}\text{.}\) As will often be the case, the first step is to rewrite in terms of a function that we already know how to differentiate, by solving for \(x\text{:}\) \(x = f(x) = y^2\text{.}\)
Using Leibniz notation first, \(dx/dy = d(y^2)/dy = 2 y\) so \(y = \sqrt{x}\) has derivative
\begin{equation*}
\frac{d \sqrt{x}}{d x} = \frac{d y}{d x} = \frac{1}{dx/dy} = \frac{1}{2 y}\text{.}
\end{equation*}
Then to get this in terms of \(x\text{,}\) use \(y = \sqrt{x}\) to get
\begin{equation*}
\frac{d \sqrt{x}}{dx} = \frac{1}{2 \sqrt{x}}\text{,}
\end{equation*}
as we have seen before.
Subsection 3.7.1 A procedure using just the Chain Rule
Though these formulas can be useful, it is in many cases easier and safer to use a strategy of
- "inverting" the equation \(y = f^{-1}(x)\) to the equation \(x= f(y)\) which involves only a function we already know how to differentiate, then
- differentiate both sides of that equation using the Chain Rule, and finally
- solve a simple equation by division.
This procedure will be used in most examples from now on, and is the basis of an important strategy introduced in Section 3.8: Implicit Differentiation. (In fact, I often prefer avoiding the memorizing of yet another formula by instead having a "procedure" or "algorithm" that break the calculation into steps each of which uses facts and methods that I already know.)
Subsection 3.7.2 The Power Rule \((x^r)' = r x^{r-1}\) for any rational number \(r\)
The result above for the square root can be extended to compute the derivative of any root function \(f(x) = \sqrt[q]{x} = x^{1/q}\) for \(q\) a natural number, and that is the main step in verifying the power rule for all rational powers. This will be done using the new strategy described above.
Firstly, since the \(q\)-th root is the inverse of the \(q\)-th power, \(y = x^{1/q}\) means that \(x = y^q\) (so the inverse function "\(\sqrt[q]{}\)" disappears). Then differentiate both sides of this equation, noting well that the variable is \(x\text{,}\) not \(y\). The Leibniz notation is safest here: On one side, \(\displaystyle\frac{dx}{dx} = 1\text{;}\) on the other side,
\begin{equation*}
\frac{d}{dx} y^q = \frac{d (y^q)}{dy} \frac{dy}{dx} = q y^{q-1} \frac{dy}{dx}
\end{equation*}
Equating these two, and inserting the formula for \(y\text{,}\)
\begin{equation*}
1 = q (x^{1/q})^{q-1} \frac{dy}{dx} = q x^{1 - 1/q} \frac{dy}{dx}
\end{equation*}
Finally, solve for the desired derivative:
\begin{equation*}
\frac{dy}{dx} = \frac{d}{dx} x^{1/q} = \frac{1}{q} x^{1/q - 1}\text{,}
\end{equation*}
confirming the Power Rule for this case of exponent \(1/q\text{.}\)
Now we can get the result for \(x^r\) where \(r=p/q\) is any rational number, by using the Chain Rule again. First, \(x^r = (x^{1/q})^p\text{,}\) \(= u^p\) with \(u = x^{1/q}\text{.}\) Then
\begin{equation*}
\frac{d}{dx}(x^r) = \frac{d u^p}{dx} = \frac{d u^p}{du} \frac{du}{dx} = \frac{d u^p}{du}\frac{d (x^{1/q})}{dx}
\end{equation*}
and we already have the power rule for each of these factors:
\begin{equation*}
\frac{d}{dx}(x^r) = p u^{p-1} \frac{1}{q}x^{1/q - 1}
\end{equation*}
Finally, get it all in terms of the variable \(x\) using \(u = x^{1/q}\) and \(p/q = r\text{:}\)
\begin{equation*}
\frac{d}{dx}(x^r) = \frac{p}{q} (x^{1/q})^{p-1} x^{1/q - 1} = \frac{p}{q} x^{p/q - 1/q + 1/q - 1} = \frac{p}{q} x^{p/q - 1} = r x^{r-1},
\end{equation*}
as advertised.
In fact we will soon be able to verify the power rule for any real power, \(x^a\text{,}\) so the above was not essential, but gives some useful examples of using the Chain Rule and of this strategy for differentiating inverses.
To complete the story of the Power Rule, we first need the derivative of the natural logarithm.
Subsection 3.7.3 Derivative of the Natural Logarithm
The Chain Rule can also be used to compute the last of the derivative of the last basic elementary functions, the logarithm. The text leaves this till Section 3.9 2 but again, I like to introduce all the elementary functions as soon as possible.
For this we again use the above strategy of solving the equation so that the inverse function temporarily "disappears" and we only have to deal with functions and operations (like composition) that we already know how to handle; this will also be very useful in the next few sections.
Let \(u = \ln x\) and solve for \(x\text{,}\) giving
\begin{equation*}
x = e^u = e^{\ln x}
\end{equation*}
Differentiating both sides with respect to \(x\) and using the Chain Rule gives
\begin{equation*}
1 = \frac{d e^u}{dx} = \frac{d e^u}{du} \frac{du}{dx} = e^u \frac{du}{dx}
\end{equation*}
Inserting \(e^u = x\) and \(u = \ln x\text{,}\) this says that \(x \frac{d}{dx}\ln x = 1\text{,}\) so dividing by \(x\) gives
\begin{equation}
\frac{d}{dx}\ln x = \frac{1}{x}, = x^{-1}\text{.}\tag{3.7.3}
\end{equation}
Example 3.7.2. The derivative of any logarithmic function: \(\displaystyle (\log_a x)' = \frac{1}{\ln a} \frac{1}{x}\).
Subsection 3.7.4 Verification of the Power Rule for all Real Powers
To differentiate \(f(x) = x^a\) for \(x > 0\) and any real \(a\text{,}\) write \(x=e^{(\ln x)}\) so that \(x^a = [e^{(\ln x)}]^a = e^{(a \ln x)}\text{.}\) Now use the Chain Rule, with
- inside function \(u=a \ln x\text{,}\) with derivative \(\displaystyle a \cdot \frac{1}{x}\text{;}\)
- outside function \(e^u\text{,}\) with derivative \(e^u, = e^{a \ln x} = x^a\text{.}\)
\begin{equation*}
\frac{d (x^a)}{dx} = \frac{d (e^u)}{dx} = \frac{d (e^u)}{du} \frac{du}{dx} = e^u \left(a \cdot \frac{1}{x}\right) = \frac{x^a \cdot a}{x} = a x^{a-1}.
\end{equation*}
Subsection 3.7.5 Derivatives of the Inverse Trigonometric Functions
We can compute the derivative of an inverse trigonometric function like \(y=\arcsin x = \sin^{-1}x\) by again using the strategy described in Subsection 3.7.1 above of first "solving for \(x\)" to hide the inverse function: writing an equation involving the original “non-inverse” function whose derivative we know and then using implicit differentiation. Here we can use
\begin{equation*}
x = \sin y
\end{equation*}
One thing we must be careful about is that no trigonometric function is invertible on its entire natural domain, so we limit the domain to make it satisfy the Horizontal Line Test. Here, we restrict the domain of \(\sin\) to \([-\pi/2,\pi/2]\) where it is increasing, cuting off the domain at the points \(x = \pm \pi/2\) beyond which it flips to being decresing. Thus \(\sin^{-1} x\) has range \([-\pi/2,\pi/2]\text{,}\) and domain \([-1,1]\text{.}\)
Implicit differentiation of \(x = \sin y\) gives \(\displaystyle 1 = \cos y \cdot \frac{dy}{dx}\text{,}\) so
\begin{equation*}
\frac{dy}{dx}= \frac{1}{\cos y}, \quad \cos y \neq 0.
\end{equation*}
But we want this as a function of \(x\text{,}\) not \(y\text{!}\)
Using the identity \(\sin^2 y + \cos^2 y = 1\) along with \(x=\sin y\text{,}\) we get \(\cos^2 y + x^2 = 1\text{,}\) so \(\cos y = \pm \sqrt{1-x^2}.\) The range of \(y\) values \([-\pi/2,\pi/2]\) ensures that \(\cos y\) is not negative, so \(\cos y = \sqrt{1-x^2}\) and
\begin{equation*}
\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1.
\end{equation*}
The restriction on the \(x\) values is to avoid division by zero, and shows that \(\arcsin x\) is not differentiable at the endpoints of its domain, due to “vertical tangents” (as with \(\sqrt{x}\) at \((0,0)\)). This happens because sine has horizontal tangents at the endpoints of its domain.
This is rather typical with inverse functions, because restricting the domain of a function to make it satisfy the Horizontal Line Test often requires ending the domain at a point where the tangent is horizontal. For example, the same happens when we cut off the domain of \(y=x^2\) at \(x=0\) to get its inverse \(\sqrt{x}\text{.}\) It also happens with the inverses of \(\cos\text{,}\) \(\sec\) and \(\csc\text{.}\)
The Derivatives of Other Inverse Trig. Functions.
Similar calculations give
\begin{gather*}
\frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx} \sec^{-1} x = \frac{1}{x\sqrt{x^2-1}}\\
\frac{d}{dx} \cos^{-1} x = -\frac{1}{\sqrt{1-x^2}} \qquad \frac{d}{dx} \csc^{-1} x = -\frac{1}{x\sqrt{x^2-1}}\\
\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2} \qquad \frac{d}{dx} \cot^{-1} x = - \frac{1}{1+x^2}
\end{gather*}
Checkpoint 3.7.3. Reciprocals vs inverses.
Checkpoint 3.7.4. Compositions vs products.
Exercises Exercises
Study Calculus Volume 1, Section 3.7 3 ; in particular Examples 61 to 67, Checkpoint items 43 to 46, and Exercises 265, 267, 269, 271, 279, and 291.
Hint for Exercise 279. One approach is to use the "equation solving" strategy of making the inverse function disappear: solve for \(\sin(y)=x^2\) and then differentiate each side of that equation.
openstax.org/books/calculus-volume-1/pages/3-7-derivatives-of-inverse-functionsopenstax.org/books/calculus-volume-1/pages/3-9-derivatives-of-exponential-and-logarithmic-functionsopenstax.org/books/calculus-volume-1/pages/3-7-derivatives-of-inverse-functions
