REMINDER The inverse relationship between the variables \(x\) and \(y\) of the function \(y=f(x)\) with an inverse function \(g\) can formally be expressed as \(x=g(y)\). This inverse function is often written as a function of \(x\) by interchanging the roles of \(x\) and \(y\), and is then denoted by \(g(x)=f^{-1}(x)\). Do not confuse this with the reciprocal \(\displaystyle \frac{1}{f(x)}\). If necessary, review the definition and properties of inverse functions in Section 3.10.

THEOREM 1 Derivative of the Inverse

Assume that \(y=f(x)\) is differentiable and one-to-one with inverse function \(x=g(y) \). If \(b\) belongs to the domain of \(g\) and \(f'(g(b)) \neq 0\), then \(g'(b)\) exists and

or, equivalently, in Leibniz notation

Proof

The first claim, that \(g\) is differentiable at \(b\) if \(f'(g(b)) \neq 0\), is verified in Appendix D (see Theorem 6). To prove Eq. (1), note that \(f(g(x)) = x\) by definition of the inverse. Differentiate both sides of this equation, and apply the Chain Rule:

Set \(x = b\) to obtain Eq. (1). To verify (2), we observe that this is obtained from (1) by noting that since \(x=g(y)\) is differentiable, we can write \(g'(y)= \frac{dx}{dy}\). Again evaluating this derivative at \(b\) and the derivative \(\frac{dy}{dx}\) on the right at \(g(b)\) yields the formula in (2).

GRAPHICAL INSIGHT

The formula for the derivative of the inverse function has a clear graphical interpretation. Consider a line \(L\) of slope \(m\) and let \(L'\) be its reflection through \(y = x\) as in Figure 3.47(A). Then the slope of \(L'\) is \(\frac{1}{m}\). Indeed, if \((a, b)\) and \((c, d)\) are any two points on \(L\), then \((b, a)\) and \((d, c)\) lie on \(L'\) and

Now recall that the graph of the inverse \(g(x)\) is obtained by reflecting the graph of \(f(x)\) through the line \(y = x\). As we see in Figure 3.47(B), the tangent line to \(y = g(x)\) at \(x = b\) is the reflection of the tangent line to \(y = f(x)\) at \(x = a\) [where \(b = f(a)\) and \(a = g(b)\)]. These tangent lines have reciprocal slopes, and thus \(g'(b) = \frac{1}{f'(a)} = \frac{1}{f'(g(b))}\), as claimed in Theorem 1.

Figure 3.47: Graphical illustration of the formula \(g'(b) = \frac{1}{f'(g(b))}\).

EXAMPLE 1 Using Formula (1)

Calculate \(g'(x)\), where \(g(x)\) is the inverse of the function \(f(x) = x^{4} + 10\) on the domain \(\{x:x\geq0\}\).

Solution Solve \(y = x^{4} + 10\) for \(x\) to obtain \(x = (y - 10)^{\frac{1}{4}}\). Thus \(g(x) = (x - 10)^{\frac{1}{4}}\). Since \(f'(x) = 4x^{3}\), we have \(f'(g(x)) = 4g(x)^{3}\), and by Eq. (1),

We obtain this same result by differentiating \(g(x) = (x - 10)^{\frac{1}{4}}\) directly.

EXAMPLE 2 Calculating \(g'(x)\;\) Without Solving for \(g(x)\)

Calculate \(g'(1)\), where \(g(x)\) is the inverse of \(f(x) = x + e^{x}\).

Solution In this case, we cannot solve for \(g(x)\) explicitly, but a formula for \(g(x)\) is not needed (Figure 3.48). All we need is the particular value \(g(1)\), which we can find by solving \(f(x) = 1\). By inspection, \(x + e^{x} = 1\) has solution \(x = 0\). Therefore, \(f(0) = 1\) and, by definition of the inverse, \(g(1) = 0\). Since \(f'(x) = 1 + e^{x}\),

Figure 3.48: Graph of \(f(x) = x + e^{x}\) and its inverse \(g(x)\).

REMINDER In Example 7 of Section 1.5, we used the right triangle in Figure 3.49 in the computation:

We note that this also works when \(\theta\) is negative, since by definition of the inverse sine, \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), and the function \(\cos \theta\) is non-negative on that interval.

THEOREM 2 Derivatives of Arcsine and Arccosine

Proof

Apply Eq. (1) with \(f(x) = \sin x\) and \(g(x) = \sin^{-1}x\). Then \(f'(x) = \cos x\), and by the equation in the margin,

Let us now derive this without use of the chain rule. \(y= \sin^{-1} x\) means that \(x= \sin y\), where \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\), and thus \(\displaystyle \frac{dx}{dy}=\cos y\) with \(\cos y \geq 0\). Therefore \(\displaystyle \cos y = \sqrt{1-\sin^2 y}\) and it follows that

\[\frac{dy}{dx}=\frac{1}{\large {\frac{dx}{dy}}}=\frac{1}{\cos y}= \frac{1}{\sqrt{1-\sin^2 y}}= \frac{1}{\sqrt{1-x^2}}\]

The computation of \(\frac{d}{dx}\cos^{-1}x\) is similar (see Exercise 37 or the next example).

EXAMPLE 3 Complementary Angles

The derivatives of \(\sin^{-1}x\) and \(\cos^{-1}x\) are equal up to a minus sign. Explain this by proving that

Solution In Figure 3.50, we have \(\theta = \sin^{-1}x\) and \(\psi = \cos^{-1}x\). These angles are complementary, so \(\theta+\psi = \frac{\pi}{2}\) as claimed. Therefore,

Figure 3.50: The angles \(\theta = \sin^{-1}x\) and \(\psi = \cos^{-1}x \) are complementary and thus sum to \(\frac{\pi}{2}\).

EXAMPLE 4

Calculate \(f'(\frac{1}{2})\), where \(f(x) = \text{arcsin}(x^{2})\).

Solution Recall that \(\text{arcsin} x\) is another notation for \(\sin^{-1}x\). By the Chain Rule,

THEOREM 3 Derivatives of Inverse Trigonometric Functions

EXAMPLE 5

Calculate \(\left.\frac{d}{dx}\csc^{-1}(e^x+1)\right|_{x=0}\).

Solution Apply the Chain Rule using the formula \(\frac{d}{dx}\csc^{-1}u = -\frac{1}{|u|\sqrt{u^2-1}}\):

We have replaced \(|e^{x} + 1|\) by \(e^{x} + 1\) because this quantity is positive. Now we have