3.2 Implicit Differentiation; Derivatives of the Inverse Trigonometric Functions

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OBJECTIVES

When you finish this section, you should be able to:

  1. Find a derivative using implicit differentiation (p. 209)
  2. Find higher-order derivatives using implicit differentiation (p. 212)
  3. Differentiate functions with rational exponents (p. 213)
  4. Find the derivative of an inverse function (p. 214)
  5. Differentiate the inverse trigonometric functions (p. 216)

So far we have differentiated only functions \(y=f( x) \) where the dependent variable \(y\) is expressed explicitly in terms of the independent variable \(x\). There are functions that are not written in the form \(y=f( x) \), but are written in the implicit form \(F( x, y) =0\). For example, \(x\) and \(y\) are related implicitly in the equations \[ xy-4=0\qquad y^{2}+3x^{2}-1=0\qquad e^{x^{2}-y^{2}}-\cos~ \left( xy\right) =0 \]

NEED TO REVIEW?

The implicit form of a function is discussed in Section P.1, p. 4.

The implicit form \(xy-4=0\) can easily be written explicitly as the function \(y=\dfrac{4}{x}\). Also, \(y^{2}+3x^{2}-1=0\) can be written explicitly as the two functions \(y_{1}=\sqrt{1-3x^{2}}\) and \(y_{2}=-\sqrt{1-3x^{2}}\). In the equation \(e^{x^{2}-y^{2}}-\cos~ \left( xy\right) =0,\) it is impossible to express \(y\) as an explicit function of \(x\). In this case, and in many others, we use the technique of implicit differentiation to find the derivative.

1 Find a Derivative Using Implicit Differentiation

The method used to differentiate an implicitly defined function is called implicit differentiation. It does not require rewriting the function explicitly, but it does require that the dependent variable \(y\) is a differentiable function of the independent variable \(x\). So throughout the section, we make the assumption that there is a differentiable function \(y=f( x) \) defined by the implicit equation. This assumption made, the method consists of differentiating both sides of the implicitly defined function with respect to \(x\) and then solving the resulting equation for \(\dfrac{dy}{dx}\).

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Finding a Derivative Using Implicit Differentiation

Find \(\dfrac{dy}{dx}\) if \(xy-4=0\).

  1. Use implicit differentiation.
  2. Solve for \(y\) and then differentiate.
  3. Verify the results of (a) and (b) are the same.

Solution(a) To differentiate implicitly, we assume \(y\) is a differentiable function of \(x\) and differentiate both sides with respect to \(x\). \[ \begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx}(xy-4) &=\dfrac{d}{dx}0 & {\color{#0066A7}{\hbox{Differentiate both sides with respect to}\ {x}.}}\\ \dfrac{d}{dx}(xy) -\dfrac{d}{dx}4 &= 0 & {\color{#0066A7}{\hbox{Use the Difference Rule.}}}\\ x\cdot \dfrac{d}{dx}y+\left(\dfrac{d}{dx}x\right) y-0 &= 0 & {\color{#0066A7}{\hbox{Use the Product Rule.}}}\\ x\dfrac{dy}{dx}+y & =0 & {\color{#0066A7}{\hbox{Simplify}.}} \\ \dfrac{dy}{dx} & = -\dfrac{y}{x} & {\color{#0066A7}{\hbox{Solve for}\, \dfrac{dy}{dx}.}} & (1) \end{array} \end{eqnarray*} \]

(b) Solve \(xy-4=0\) for \(y\), obtaining \(y=\dfrac{4}{x}=4x^{-1}\). Then \[ \begin{equation*} \dfrac{dy}{dx}=\dfrac{d}{dx}(4x^{-1}) =-4x^{-2}=-\dfrac{4}{x^{2}}\tag{2} \end{equation*} \]

(c) At first glance, the results in (1) and (2) appear to be different. However, since \(xy-4=0,\) or equivalently, \(y=\dfrac{4}{x},\) the result from (1) becomes \[ \begin{eqnarray*} && \dfrac{dy}{dx}\underset{\underset{\color{#0066A7}{(1)}}{\color{#0066A7}{\uparrow}}}{=}-\dfrac{y}{x}\underset{\underset{\color{#0066A7}{y=\tfrac{4}{x}}}{\color{#0066A7}{\uparrow}}}{=} -{\dfrac{\dfrac{4}{x}}{x}}=-\dfrac{4}{x^{2}} \end{eqnarray*} \] which is the same as (2).

In most instances, we will not know the explicit form of the function (as we did in Example 1) and so we will leave the derivative \(\dfrac{dy}{dx}\) expressed in terms of \(x\) and \(y\) [as in (1)].

NOW WORK

Problem 17.

The Power Rule for Functions is \[ \dfrac{d}{dx}[f(x)] ^{n}=n[f(x)] ^{n-1}f^\prime (x) \] where \(n\) is an integer. If \(y=f( x)\), it takes the form \[\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{d}{dx}y^{n}=ny^{n-1}\dfrac{dy}{dx}} \]

This is convenient notation to use with implicit differentiation when \(y^n\) appears. \[ \dfrac{d}{dx}y\underset{\underset{\color{#0066A7}{\hbox{\(n\)=1}}}{\color{#0066A7}{\uparrow}}}{=}1\cdot \dfrac{dy}{dx}=\dfrac{dy}{dx}\quad\quad \dfrac{d}{dx}y^{2}\underset{\underset{\color{#0066A7}{\hbox{\(n\)=2}}}{\color{#0066A7}{\uparrow}}}{=}2y\dfrac{dy}{dx}\quad\quad \dfrac{d}{dx}y^{3}\underset{\underset{\color{#0066A7}{\hbox{\(n=3\)}}}{\color{#0066A7}{\uparrow}}}{=}3y^{2}\dfrac{dy}{dx} \]

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To differentiate an implicit function:

  • Assume that \(y\) is a differentiable function of \(x\).
  • Differentiate both sides of the equation with respect to \(x\).
  • Solve the resulting equation for \(y^\prime =\dfrac{dy}{dx}.\)

Finding a Derivative Using Implicit Differentiation

Find \(\dfrac{dy}{dx}\) if \(3x^{2}+4y^{2}=2x\).

Solution We assume that \(y\) is a differentiable function of \(x\) and differentiate both sides with respect to \(x\). \[ \begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx} ( 3x^{2}+4y^{2}) &= \dfrac{d}{dx}(2x) & {\color{#0066A7}{\hbox{Differentiate both sides with respect to}\; x.}}\\ \dfrac{d}{dx} (3x^{2}) +\dfrac{d}{dx}( 4y^{2}) &= 2 & {\color{#0066A7}{\hbox{Sum Rule}.}} \\ 3\dfrac{d}{dx}x^{2}+4\dfrac{d}{dx}y^{2} &=2 &{\color{#0066A7}{\hbox{Constant Multiple Rule}.}}\\ 6x+4\!\left( 2y\dfrac{dy}{dx}\right) &= 2 &{\color{#0066A7}{\dfrac{d}{dx}{y}^{{ 2}}{=2y} \dfrac{dy}{dx}}}\\ 6x+8y\dfrac{dy}{dx} &= {2} & {\color{#0066A7}{\hbox{Simplify.}}}\\ \dfrac{dy}{dx} &= \dfrac{2-6x}{8y}=\dfrac{1-3x}{4y} & {\color{#0066A7}{\hbox{Solve for}\; \dfrac{dy}{dx}.}} \end{array} \end{eqnarray*} \] provided \(y\ne 0\).

Notice in Example 2 that \(\dfrac{dy}{dx}\) is expressed in terms of \(x\) and \(y\).

NOW WORK

Problem 15.

Using Implicit Differentiation to Find an Equation of a Tangent Line

Find an equation of the tangent line to the graph of the ellipse \(3x^{2}+4y^{2}=2x\) at the point \(\left( \dfrac{1}{2},-\dfrac{1}{4}\right)\).

Solution First we find the slope of the tangent line. We use the result from Example 2, and evaluate \(\dfrac{dy}{dx}=\dfrac{1-3x}{4y}\) at \(\left( \dfrac{1}{2},-\dfrac{1}{4}\right) \). \[ \begin{eqnarray*} && \dfrac{dy}{dx}=\dfrac{1-3x}{4y} \underset{\underset{\color{#0066A7}{x=\tfrac{1}{2},y=-\dfrac{1}{4}}}{\color{#0066A7}{{\left\uparrow{\vphantom{\vrule width0pc height12.5pt depth0pt}}\right.}}}}{=} \dfrac{1-3\cdot \dfrac{1}{2}}{4\,{\cdot}\,\!\left(-\dfrac{1}{4}\right) }=\dfrac{1}{2}\\ \end{eqnarray*} \]

The slope of the tangent line to the graph of \(3x^{2}+4y^{2}=2x\) at the point \(\left( \dfrac{1}{2},-\dfrac{1}{4}\right) \) is \(\dfrac{1}{2}\). An equation of the tangent line is \[ \begin{eqnarray*} y+\dfrac{1}{4} &=&\dfrac{1}{2}\!\left( x-\dfrac{1}{2}\right) \\ y &=&\dfrac{1}{2}x-\dfrac{1}{2} \end{eqnarray*} \]

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Figure 2 shows the graph of the ellipse \(3x^{2}+4y^{2}=2x\) and the graph of the tangent line \(y=\dfrac{1}{2}x-\dfrac{1}{2}\) at the point \(\left( \dfrac{1}{2},-\dfrac{1}{4}\right) \).

Figure 2

NOW WORK

Problem 63.

Using Implicit Differentiation

  1. Find \(y^\prime \) if \(e^{y}\cos x=x+1\).
  2. Find an equation of the tangent line to the graph at the point \((0,0) \).

Solution(a) We use implicit differentiation: \[ \begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx}( e^{y}\cos x) &= \dfrac{d}{dx}(x+1) \\ e^{y}\cdot \dfrac{d}{dx}( \cos x) +\left( \dfrac{d}{dx} e^{y}\right) \cdot \cos x &= 1 & {\color{#0066A7}{\hbox{Use the Product Rule.}}} \\ e^{y}( -\sin x) +e^{y}y^\prime \cdot \cos x &= 1 \\ ( e^{y}\cos x) y^\prime &= 1+e^{y}\sin x \\ y^\prime &= \dfrac{1+e^{y}\sin x}{e^{y}\cos x} \end{array} \end{eqnarray*} \]

(b) At the point \((0,0) \), the derivative \(y^\prime \) is \(\dfrac{1+e^{0}\sin 0}{e^{0}\cos 0}=\dfrac{1+1\cdot 0}{1\cdot 1}=1\), so the slope of the tangent line to the graph at \((0,0) \) is \(1\). An equation of the tangent line to the graph at the point \((0,0)\) is \(y=x.\)

Figure 3

The graph of \(e^{y}\cos x=x+1\) and the tangent line to the graph at \((0,0)\) are shown in Figure 3.

2 Find Higher-Order Derivatives Using Implicit Differentiation

Implicit differentiation can be used to find higher-order derivatives.

Finding Higher-Order Derivatives

Use implicit differentiation to find \(y^\prime \) and \(y^{\prime\prime} \) if \(y^{2}-x^{2}=5\). Express \(y^{\prime\prime} \) in terms of \(x\) and \(y\).

Solution First, we assume there is a differentiable function \(y=f( x) \) that satisfies \(y^{2}-x^{2}=5\). Now we find \(y^\prime \).



provided \(y≠ 0\).

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Equations (3) and (4) both involve \(y^\prime \). Either one can be used to find \(y^{\prime\prime} .\) We use (3) because it avoids differentiating a quotient. \[ \begin{eqnarray*} \dfrac{d}{dx} ( 2yy^\prime -2x) &=&\dfrac{d}{dx}0 \nonumber\\ \dfrac{d}{dx} ( yy^\prime ) -\dfrac{d}{dx}x &=&0 \nonumber\\ y\cdot \dfrac{d}{dx}y^\prime +\left( \dfrac{d}{dx}y\right) y^\prime -1 &=&0 \nonumber\\ yy^{\prime\prime} +( y^\prime ) ^{2}-1 &=&0 \nonumber\\ y^{\prime\prime} &=&\dfrac{1-( y^\prime ) ^{2}}{y}\tag{5} \end{eqnarray*} \] provided \(y≠ 0\). To express \(y^{\prime\prime} \) in terms of \(x\) and \(y,\) we use (4) and substitute for \(y^\prime \) in (5).

NOW WORK

Problem 59.

3 Differentiate Functions with Rational Exponents

In Section 2.4, we showed that the Simple Power Rule \[ \dfrac{d}{dx}x^{n}=nx^{n-1} \] is true if the exponent \(n\) is any integer. We use implicit differentiation to extend this result for rational exponents.

THEOREM Power Rule for Rational Exponents

If \(y=x^{p/q}\), where \(\dfrac{p}{q}\) is a rational number, then \[ \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{y^\prime =\dfrac{d}{dx}x^{p/q}=\dfrac{p}{q}\cdot x^{\left(p/q\right) -1}}\tag{6} \end{equation*} \] provided that \(x^{p/q}\) and \(x^{(p/q) -1}\) are defined.

Proof

We begin with the function \(y=x^{p/q},\) where \(p\) and \(q>0\) are integers. Now, we raise both sides of the equation to the power \(q\) to obtain \[ y^{q}=x^{p} \]

This is the implicit form of the function \(y=x^{p/q}.\) Assuming \(y\) is differentiable*, we can differentiate implicitly, obtaining \[ \begin{eqnarray*} \frac{d}{dx}y^q &=& \frac{d}{dx} x^p\\ qy^{q-1}y^\prime &=& px^{p-1} \end{eqnarray*} \]

*In Problem 113, you are asked to show that \(y=x^{p/q}\) is differentiable.

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Now solve for \(y^\prime\). \[ \begin{eqnarray*} && y^\prime =\dfrac{px^{p-1}}{qy^{q-1}}\underset{\underset{\color{#0066A7}{y=x^{p/q}}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{p}{q}\cdot \dfrac{x^{p-1}}{( x^{p/q}) ^{q-1}}=\dfrac{p}{q} \cdot \dfrac{x^{p-1}}{x^{p-(p/q) }}=\dfrac{p}{q}\cdot x^{p-1-[ p-(p/q)] }=\dfrac{p}{q}\cdot x^{(p/q)-1}\\ \end{eqnarray*} \]

Differentiating Functions with Rational Exponents

  1. \(\dfrac{d}{dx}\sqrt{x}=\dfrac{d}{dx}x^{1/2}=\dfrac{1}{2} x^{( 1/2) -1}=\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2x^{1/2}}=\dfrac{1}{2 \sqrt{x}}=\dfrac{\sqrt{x}}{2x} \)
  2. \(\dfrac{d}{du}\sqrt[3]{u}=\dfrac{d}{du}u^{1/3}=\dfrac{1}{3} u^{-2/3}=\dfrac{1}{3\sqrt[3]{u^{2}}}=\dfrac{\sqrt[3]{u}}{3u}\)
  3. \(\dfrac{d}{dx}x^{5/2}=\dfrac{5}{2}x^{3/2}\)
  4. \(\dfrac{d}{ds}s^{-3/2}=-\dfrac{3}{2}s^{-5/2}=-\dfrac{3}{2s^{5/2}}\)

NOW WORK

Problem 31.

THEOREM Power Rule for Functions

If \(u\) is a differentiable function of \(x\) and \(r\) is a rational number, then \[\bbox[5px, border:1px solid black, #F9F7ED] {\dfrac{d}{dx}[u(x)]^{r}=r[u(x)] ^{r-1}u^\prime (x)} \] provided \(u^{r}\) and \(u^{r-1}\) are defined.

Differentiating Functions Using the Power Rule

  1. \(\dfrac{d}{ds}(s^{3}-2s+1) ^{5/3}=\dfrac{5}{3} (s^{3}-2s+1) ^{2/3}\dfrac{d}{ds}(s^{3}-2s+1) = \dfrac{5}{3}(s^{3}-2s+1) ^{2/3}(3s^{2}-2) \)
  2. \(\begin{align*} \dfrac{d}{dx}\sqrt[3]{x^{4}-3x+5} &=\dfrac{d}{dx}(x^{4}-3x+5) ^{1/3}=\dfrac{1}{3}(x^{4}-3x+5) ^{-2/3}\dfrac{d}{dx}(x^{4}-3x+5) \\&= \dfrac{4x^{3}-3}{3(x^{4}-3x+5)^{2/3}} \end{align*}\)
  3. \(\begin{align*} \dfrac{d}{d\theta }[\tan (3\theta ) ] ^{-3/4}&=-\dfrac{3}{4}[\tan (3\theta ) ] ^{-7/4} \dfrac{d}{d\theta }\tan (3\theta ) = -\dfrac{3}{4}[\tan (3\theta ) ] ^{-7/4}\cdot \sec ^{2}(3\theta ) \cdot 3 \\&= -\dfrac{9\sec^2 (3\theta ) }{4[\tan (3\theta ) ] ^{7/4}} \end{align*}\)

NOW WORK

Problem 39.

NEED TO REVIEW?

Inverse functions are discussed in Section P.4, pp. 32-37.

4 Find the Derivative of an Inverse Function

Suppose \(f\) is a function and \(g\) is its inverse function. Then \[ g( f( x) ) =x \] for all \(x\) in the domain of \(f.\)

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If both \(f\) and \(g\) are differentiable, we can differentiate both sides with respect to \(x\) using the Chain Rule. Then \[ \begin{eqnarray*} && \dfrac{d}{dx}[g(f(x))] \underset{\underset{\underset{\underset{\color{#0066A7}{\hbox{on the left}}}{\color{#0066A7}{\hbox{Chain Rule}}}}{\color{#0066A7}{\hbox{Use the}}}}{\color{#0066A7}{\uparrow}}}{=} g^\prime (f(x)) \cdot f^\prime ( x) \underset{\underset{\underset{\color{#0066A7}{\hbox{on the right}}}{\color{#0066A7}{\hbox{\(\tfrac{d}{dx}x=1\)}}}}{\color{#0066A7}{\uparrow}}}{=} 1\\ \\ \end{eqnarray*} \]

Since the product of the derivatives is never \(0\), each function has a nonzero derivative on its domain.

Conversely, if a one-to-one function has a nonzero derivative, then its inverse function also has a nonzero derivative as stated in the following theorem.

THEOREM Derivative of an Inverse Function

Let \(y=f( x) \) and \(x=g( y) \) be inverse functions. Suppose \(f\) is differentiable on an open interval containing \(x_{0}\) and \(y_{0}=f( x_{0}) .\) If \(f^\prime ( x_{0}) ≠ 0\), then \(g\) is differentiable at \(y_{0}=f( x_{0}) \) and \[ \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{d}{dy}g(y_{0}) =g^\prime (y_{0}) =\dfrac{1}{f^\prime (x_{0})}}\tag{7} \end{equation*} \] where the notation \(\dfrac{d}{dy}g( y_{0})=g^\prime(y_{0}) \) means the value of \(\dfrac{d}{dy}g( y) \) at \(y_{0},\) and the notation \(f^\prime ( x_{0}) \) means the value of \(f^\prime ( x) \) at \(x_{0}\).

In Leibniz notation, formula (7) has the simple form \[\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}} \]

We have two comments to make about this theorem, which is proved in Appendix B:

  • If \(y_{0}=f( x_{0}) \) and \(f^\prime ( x_{0}) ≠ 0\) exists, then \(\dfrac{d}{dy}g( y) \) exists at \(y_{0}.\)
  • It gives formula \(( 7) \) for finding \(\dfrac{d}{dy}g(y) \) at \(y_{0}\) without knowing a formula for \(g=f^{-1}\), provided we can find \(x_{0}\) and \(f^\prime ( x_{0}) .\)

Finding the Derivative of an Inverse Function

The function \(f( x) =x^{5}+x\) has an inverse function \(g\). Find \(g^\prime (2) \).

Solution Using (7) with \(y_{0}=2\), we get \[ g^\prime (2) =\dfrac{1}{f^\prime (x_{0})}\qquad \hbox{where}\ 2=f( x_{0}) \]

A solution of the equation \[ f( x_{0}) =x_{0}^{5}+x_{0}=2 \] is \(x_{0}=1\). Since \(f^\prime ( x) =5x^{4}+1\), then \( f^\prime (x_{0})= f^\prime (1) =6\) and \[ g^\prime (2) =\dfrac{1}{f^\prime (1) }=\dfrac{1}{6} \]

Observe in Example 8 that the derivative of the inverse function \(g\) at \(2\) was evaluated without actually knowing a formula for \(g\).

NOW WORK

Problem 71.

5 Differentiate the Inverse Trigonometric Functions

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Table 1 lists the inverse trigonometric functions and their domains.

NEED TO REVIEW?

Inverse trigonometric functions are discussed in Section P.7, pp. 58-61.

Table 1: TABLE 1
\(f\) Restricted Domain \(f^{-1}\) Domain
\(f(x) =\sin x\) \(\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right] \) \(f^{-1}( x) =\sin ^{-1}x\) \([-1,1] \)
\(f(x) =\cos x\) \([0,\pi] \) \(f^{-1}( x) =\cos ^{-1}x\) \([-1,1] \)
\(f(x) =\tan x\) \(\left( -\dfrac{\pi }{2},\dfrac{\pi }{2}\right) \) \(f^{-1}( x) =\tan ^{-1}x\) \((-\infty ,\infty) \)
\(f(x) =\csc x\) \(\left(-\pi, -\dfrac{\pi}{2}\right]\cup \left(0, \dfrac{\pi}{2}\right]\) \( f^{-1}( x) =\csc ^{-1}x\) \(|x| ≥ 1\)
\(f(x) =\sec x\) \(\left[0, \dfrac{\pi}{2}\right)\cup \left[\pi, \dfrac{3\pi}{2}\right)\) \( f^{-1}(x) =\sec ^{-1}x\) \(|x| ≥ 1\)
\(f(x) =\cot x\) \((0,\;\pi) \) \(f^{-1}( x) =\cot ^{-1}x\) \((-\infty ,\infty) \)

To find the derivative of \(y=\sin ^{-1}x,\) \(-1≤ x≤ 1,\) \(-\dfrac{\pi }{2}≤ y≤ \dfrac{\pi }{2},\) we write \(\sin y=x\) and differentiate implicitly with respect to \(x\). \[ \begin{eqnarray*} \dfrac{d}{dx}\sin y &=&\dfrac{d}{dx}x \\ \cos y\cdot \dfrac{dy}{dx} &=&1 \\ \dfrac{dy}{dx} &=&\dfrac{1}{\cos y} \end{eqnarray*} \] provided \(\cos y≠ 0\). Since \(\cos y=0\) if \(y=-\dfrac{\pi }{2}\) or \(y=\dfrac{\pi }{2}\), we exclude these values. Then \(\dfrac{d}{dx}\sin^{-1} x=\dfrac{1}{\cos y}\), \(-\dfrac{\pi }{2}< y< \dfrac{\pi }{2}.\) Now, if \(-\dfrac{\pi }{2}< y< \dfrac{\pi }{2}\), then \(\cos y> 0\). Using a Pythagorean identity, we have \[ \begin{eqnarray*} \cos ^{2}y &=&1-\sin ^{2}y \\ \cos y &\underset{\underset{\color{#0066A7}{\hbox{\(\cos y>0\)}}}{\color{#0066A7}{\uparrow}}}{=}& \sqrt{1-\sin ^{2}y} \underset{\underset{\color{#0066A7}{\hbox{\(\sin y=x\)}}}{\color{#0066A7}{\uparrow}}}{=} {\sqrt{1-x^{2}}}\\ \end{eqnarray*} \]

THEOREM Derivative of \(y=\sin ^{-1}x\)

The derivative of the inverse sine function \(y=\sin ^{-1}x\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime = \dfrac{d}{dx}\sin ^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}} \qquad -1< x< 1} \]

Using the Chain Rule with the Inverse Sine Function

Find \(y^\prime \) if:
(a) \(y=\sin ^{-1}(4x^{2}) \) \( \quad \) (b) \(y=e^{\sin ^{-1}x}\)

Solution(a) If \(y=\sin ^{-1}u\) and \(u=4x^{2}\), then \(\dfrac{dy}{du}=\dfrac{1}{\sqrt{1-u^{2}}}\) and \(\dfrac{du}{dx}=8x.\) By the Chain Rule, \[ \begin{eqnarray*} && y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\left( \dfrac{1}{ \sqrt{1-u^{2}}}\right) (8x) = \dfrac{8x}{\sqrt{1-16x^{4}}}\qquad {\color{#0066A7}{\hbox{\(u=4x^{2}\)}}} \end{eqnarray*} \]

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(b) If \(y=e^{u}\) and \(u=\sin ^{-1}x\), then \(\dfrac{dy}{du}=e^{u}\) and \(\dfrac{du}{dx}=\dfrac{1}{\sqrt{1-x^{2}}}\). By the Chain Rule, \[ \begin{eqnarray*} && y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=e^{u}\cdot \dfrac{ 1}{\sqrt{1-x^{2}}} \underset{\underset{\color{#0066A7}{\hbox{\(u=\sin ^{-1}x\)}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{ e^{\sin ^{-1}x}}{\sqrt{1-x^{2}}} \\ \end{eqnarray*} \]

NOW WORK

Problem 51.

To derive a formula for the derivative of \(y\,{=}\,\tan^{-1}x,\) \(-\infty\,{< }\,x\,{< }\,\infty ,\) \(-\dfrac{\pi }{2}\,{< }\,y\,{<}\,\dfrac{\pi }{2},\) we write \(\tan y=x\) and differentiate with respect to \(x\). Then \[ \begin{eqnarray*} \dfrac{d}{dx}\tan y &=&\dfrac{d}{dx}x \\ \sec ^{2}y\dfrac{dy}{dx} &=&1 \\ \dfrac{dy}{dx} &=&\dfrac{1}{\sec ^{2}y} \end{eqnarray*} \]

Since \(-\dfrac{\pi }{2}< y< \dfrac{\pi }{2},\) then \(\sec y≠ 0.\)

Now we use the Pythagorean identity \(\sec ^{2}y=1+\tan ^{2}y,\) and substitute \(x=\tan y\). Then \[ y^\prime =\dfrac{d}{dx}\tan ^{-1}x=\dfrac{1}{1+x^{2}} \]

THEOREM Derivative of \({y=\tan ^{-1}x}\)

The derivative of the inverse tangent function \(y=\tan ^{-1}x\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\tan ^{-1}x=\dfrac{1}{1+x^{2}}} \]

The domain of \(y^\prime \) is all real numbers.

Using the Chain Rule with the Inverse Tangent Function

Find \(y^\prime \) if:
(a) \(y=\tan ^{-1}( 4x)\) \( \quad \) (b) \(y=\sin ( \tan ^{-1}x)\)

Solution(a) Let \(y=\tan ^{-1}u\) and \(u=4x.\) Then \(\dfrac{dy}{du}=\dfrac{1}{1+u^{2}}\) and \(\dfrac{du}{dx}=4.\) By the Chain Rule, \[ y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\dfrac{1}{1+u^{2}} \cdot 4=\dfrac{4}{1+16x^{2}} \]

(b) Let \(y=\sin u\) and \(u=\tan ^{-1}x\). Then \(\dfrac{dy}{du}=\cos u\) and \(\dfrac{du}{dx}=\dfrac{1}{1+x^{2}}\). By the Chain Rule, \[ y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\cos u\cdot \dfrac{1}{1+x^{2}}=\dfrac{\cos ( \tan ^{-1}x) }{1+x^{2}} \]

NOW WORK

Problem 55.

Finally to find the derivative of \(y=\sec ^{-1}x,\) \(\vert x\vert ≥ 1,\) \(0≤ y< \dfrac{\pi }{2}\) or \(\pi ≤ y< \dfrac{3\pi }{2},\) we write \(x=\sec y\) and use implicit differentiation. \[ \begin{eqnarray*} \dfrac{d}{dx}x &=&\dfrac{d}{dx}\sec y \\ 1 &=&\sec y\tan y\dfrac{dy}{dx} \end{eqnarray*} \]

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We can solve for \(\dfrac{dy}{dx}\) provided \(\sec y≠ 0\) and \(\tan y≠ 0,\) or equivalently, \(y≠ 0\) and \(y≠ \pi.\) That is, \[ \dfrac{dy}{dx}=\dfrac{1}{\sec y\tan y},\qquad \hbox{provided}\; 0< y< \dfrac{\pi }{2} \ \hbox{or}\ \pi < y< \dfrac{3\pi }{2}. \]

With these restrictions on \(y,\) \(\tan y>0\).* Now we use the Pythagorean identity \(1+\tan ^{2}y=\sec ^{2}y\). Then \(\tan y = \sqrt{\sec^2 y-1}\) and \[ \begin{eqnarray*} && \dfrac{1}{\sec y\tan y}=\dfrac{1}{\sec y\sqrt{\sec ^{2}y-1}} \underset{\underset{\color{#0066A7}{\hbox{\(\sec y=x\)}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{1}{x\sqrt{x^{2}-1}}\\ \end{eqnarray*} \]

THEOREM Derivative of \({y=\sec ^{-1}x}\)

The derivative of the inverse secant function \(y=\sec ^{-1}x\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\sec ^{-1}x=\dfrac{1}{x\sqrt{x^{2}-1}}\qquad \vert x\vert > 1 } \]

Notice that \(y=\sec ^{-1}x\) is not differentiable when \(x=\pm 1\). In fact, as Figure 4 shows, at the points \(\left( -1,\pi \right) \) and \((1,0)\), the graph of \(y=\sec ^{-1}x\) has vertical tangent lines.

Figure 4 \(y = \sec^{-1} x, \vert x\vert ≥ 1,\) \(0 ≤ y < \frac{\pi}{2}\, \hbox{or}\, \pi ≤ y < \frac{3\pi}{2}\)

The formulas for the derivatives of the three remaining inverse trigonometric functions can be obtained using the following identities:

  • Since \(\cos ^{-1}x=\dfrac{\pi }{2}-\sin ^{-1}x\), then \(\dfrac{d}{dx} \cos ^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}\), where \(\vert x\vert\,{< }\,1\).
  • Since \(\cot ^{-1}x=\dfrac{\pi }{2}-\tan ^{-1}x,\) then \(\dfrac{d}{dx} \cot ^{-1}x=-\dfrac{1}{1+x^{2}}\), where \(-\infty\,{< }\,x\,{< }\,\infty \).
  • Since \(\csc ^{-1}x=\dfrac{\pi }{2}-\sec ^{-1}x,\) then \(\dfrac{d}{dx} \csc ^{-1}x=-\dfrac{1}{x\sqrt{x^{2}-1}}\), where \(\vert x\vert\,{>}\,1\).

*The restricted domain of \(y=\sec x\) was chosen as \(\left\{x|0≤ x< \dfrac{\pi }{2}\hbox{ or } \pi ≤ x< \dfrac{3\pi }{2}\right\}\) so that \(\tan x > 0\).