2.5 The Derivative of the Trigonometric Functions

OBJECTIVES

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

  1. Differentiate trigonometric functions (p. 185)

1 Differentiate Trigonometric Functions

NEED TO REVIEW?

The trigonometric functions are discussed in Section P.6, pp. 49-56.

To find the derivatives of \(y=sin x\) and \(y=cos x\), we use the limits \[ \lim\limits_{\theta \rightarrow 0}\dfrac{\sin \theta }{\theta }=1\qquad \hbox{and}\qquad \lim\limits_{\theta \rightarrow 0} \dfrac{\cos \theta -1}{\theta }=0 \]

that were established in Section 1.4.

186

THEOREM Derivative of \(y\) = sin \(x\)

The derivative of \(y\) = sin \(x\) is \(y^\prime =\cos x\). That is, \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt] {y^\prime = \dfrac{d}{dx}\sin x=\cos x}} \]

NEED TO REVIEW?

Basic trigonometric identities are discussed in Appendix A.4, pp. A-32 to A-35.

Proof

\begin{array}{lcll} y^\prime &=&\lim\limits_{h\rightarrow 0}\dfrac{\sin(x+h) -\sin x }{h} & {\color{#0066A7}{\hbox{The definition of a derivative.}}} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{\sin x\cos h+\sin h\cos x-\sin x}{h} & {\color{#0066A7}{\hbox{\(\sin (A+B) =\sin A\,\cos B+\sin B\,\cos A.\)}}} \\ &=&\lim\limits_{h\rightarrow 0}\left[ \dfrac{\sin x\cos h-\sin x}{h}+\dfrac{\sin h\cos x}{h}\right] & {\color{#0066A7}{\hbox{Rearrange terms.}}} \\ &=&\lim\limits_{h\rightarrow 0}\left[ \sin x\cdot \dfrac{\cos h-1}{h}+\dfrac{\sin h}{h}\cdot \cos x\right] & {\color{#0066A7}{\hbox{Factor.}}} \\ &=&\left[\lim\limits_{h\rightarrow 0}\sin x\right]\! \left[\lim\limits_{h\rightarrow 0} \dfrac{\cos h-1}{h}\right] +\left[\lim\limits_{h\rightarrow 0}\cos x\right]\!\left[ \lim\limits_{h\rightarrow 0}\dfrac{\sin h}{h}\right] & {\color{#0066A7}{\hbox{Use properties of limits.}}}\\ &=&\sin x\cdot 0+\cos x\cdot 1=\cos x & {\color{#0066A7}{\hbox{\(\lim\limits_{\theta \rightarrow 0}\dfrac{\cos \theta -1}{\theta } =0 ; \quad \lim\limits_{\theta \rightarrow 0}\dfrac{\sin \theta}{\theta } =1.\)}}} \end{array}

The geometry of the derivative \(\dfrac{d}{dx}\sin x=\cos x\) is shown in Figure 26. On the graph of \(f(x) =\sin x\), the horizontal tangents are marked as well as the tangent lines that have slopes of 1 and −1. The derivative function is plotted on the second graph and those points are connected with a smooth curve.

Figure 26

Differentiating the Sine Function

Find \(y^\prime\) if:

  1. \(y=x+4\sin x\)
  2. \(y=x^{2}\sin x\)
  3. \(y=\dfrac{\sin x}{x}\)
  4. \(y=e^{x}\sin x\)

Solution

  1. \(y^\prime =\dfrac{d}{dx}(x+4\sin x)\) = \(\dfrac{d}{dx}x+\dfrac{d}{dx}(4 \sin x)\) = 1+4 \(\dfrac{d}{dx} \sin x\) = 1 + 4 \(\cos x\)
  2. \(y^\prime = \dfrac{d}{dx}(x^{2} \sin x)\) = \(x^{2} \left[\dfrac{d}{dx} \sin x\right] +\left[\dfrac{d}{dx}x^{2}\right] \sin x\) = \(x^{2}\cos x+2x\sin x\)

    187

  3. \(y^\prime =\dfrac{d}{dx}\left(\dfrac{\sin x}{x}\right)\) = \(\dfrac{\left[\dfrac{d}{dx}\sin x\right] \cdot x-\sin x\cdot \left[ \dfrac{d }{dx}x\right]}{x^{2}}=\dfrac{x\cos x-\sin x}{x^{2}}\)
  4. \[ \begin{eqnarray*} y^\prime &=& \dfrac{d}{dx}(e^{x}\sin x) = (e^{x} \dfrac{d}{dx}\sin x + \left(\dfrac{d}{dx}e^{x}\right) \sin x = e^{x}\cos x+e^{x}\sin x\\ &=& e^{x} (\cos x+\sin x) \end{eqnarray*} \]

NOW WORK

Problems 5 and 29.

THEOREM Derivative of \(y\)= cos \(x\)

The derivative of \(y= \cos x\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\cos x=-\sin x } \]

You are asked to prove this in Problem 75.

Differentiating Trigonometric Functions

Find the derivative of each function:

  1. \(f(x) =x^{2}\cos x\)
  2. \(g(\theta ) =\dfrac{\cos \theta }{1-\sin \theta }\)
  3. \(F(t) =\dfrac{e^{t}}{\cos t}\)

Solution

  1. \[ \begin{eqnarray*} f^\prime (x) &=& \frac{d}{dx} (x^{2}\cos x) = x^{2}\frac{d}{dx}\cos x + (\frac{d}{dx} x^{2}) (\cos x)\\ &=& x^{2}(-\sin x) + 2x\cos x = 2x\cos x - x^{2}\sin x \end{eqnarray*} \]
  2. \[ \begin{eqnarray*} g^\prime (\theta) &=& \frac{d}{d\theta} (\frac{\cos\theta}{1-\sin\theta}) = \frac{(\frac{d}{d\theta}\cos\theta) (1-\sin\theta) -(\cos\theta) [\frac{d}{d\theta}(1-\sin\theta) ]}{(1-\sin\theta)^{2}}\\ &=& \frac{-\sin\theta (1-\sin\theta) -\cos\theta (-\cos\theta)}{(1-\sin\theta)^{2}} =\frac{-\sin\theta +\sin^{2}\theta +\cos^{2}\theta}{(1-\sin\theta)^{2}}\\ &=&\frac{-\sin\theta+1}{(1-\sin\theta)^{2}}=\frac{1}{1-\sin\theta} \end{eqnarray*} \]
  3. \[ \begin{eqnarray*} F\prime(t) &=&\frac{d}{dt}(\frac{e^{t}}{\cos t}) = \frac{(\frac{d}{dt}e^{t}) (\cos t) -e^{t}(\frac{d}{dt}\cos t)}{\cos^{2}t}=\frac{e^{t}\cos t-e^{t}(-\sin t) }{\cos ^{2}t}\\ &=&\frac{e^{t}(\cos t+\sin t)}{\cos^{2}t} \end{eqnarray*} \]

NOW WORK

Problem 13.

Identifying Horizontal Tangent Lines

Find all points on the graph of \(f(x) =x+\sin x\) where the tangent line is horizontal.

Solution Since tangent lines are horizontal at points on the graph of \(f\) where \(f^\prime (x) =0,\) we begin by finding \(f^\prime\): \[f^\prime (x) =1+\cos x\] Now we solve the equation \(f^\prime (x) =0.\) \begin{array}{rcl@{\qquad}l} f^\prime (x) &=& 1+\cos x=0 \\[0pt] \cos x &=& -1 \\[0pt] x &=& (2k+1) \pi &\hbox{where }{k}\hbox{ is an integer} \end{array}

188

Figure 27 \(f(x)=x+\sin x\)

The graph of \(f( x) =x+\sin x\) has a horizontal tangent line at each of the points \(( ( 2k+1) \pi , ( 2k+1) \pi )\), \(k\) an integer. See Figure 27 for the graph of \(f\) with the horizontal tangent lines marked. Notice that each of the points with a horizontal tangent line lies on the line \(y=x.\)

NOW WORK

Problem 57.

The derivatives of the remaining four trigonometric functions are obtained using trigonometric identities and basic derivative rules. We establish the formula for the derivative of \(y=\tan x\) in Example 4. You are asked to prove formulas for the derivative of the secant function, the cosecant function, and the cotangent function in the exercises. (See Problems 76–78.)

Differentiating \(y\)= tan\(x\)

Show that the derivative of \(y\) = tan \(x\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\tan x=\sec ^{2}x} \]

Solution \[ \begin{eqnarray*} y^\prime &=&\dfrac{d}{dx}\tan x\underset{\underset{\color{#0066A7}{\hbox{Identity}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{d}{dx}\dfrac{\sin x}{\cos x}\underset{\underset{\color{#0066A7}{\hbox{Quotient Rule}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{\left[ \dfrac{d}{dx}\sin x\right] \cos x-\sin x\left[ \dfrac{d}{dx}\cos x\right] }{\cos ^{2}x} \\ &=&\dfrac{\cos x\cdot \cos x-\sin x\cdot (-\sin x)}{\cos ^{2}x}=\dfrac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=\dfrac{1}{\cos ^{2}x}=\sec ^{2}x \end{eqnarray*} \]

NOW WORK

Problem 15.

Table 4 lists the derivatives of the six trigonometric functions along with the domain of each derivative.

Table 4: TABLE 4
Derivative Function Domain of the Derivative Function
\(\dfrac{d}{dx}\sin x=\cos x\) \(( -\infty ,\infty ) \)
\(\dfrac{d}{dx}\cos x=-\sin x\) \(( -\infty ,\infty ) \)
\(\dfrac{d}{dx}\tan x=\sec ^{2}x\) \(\left\{ x|x\neq \dfrac{2k+1}{2}\pi, k\hbox{ an integer}\right\} \)
\(\dfrac{d}{dx}\cot x=-\csc ^{2}x\) \(\{ x|x\neq k\pi, k \hbox{ an integer}\} \)
\(\dfrac{d}{dx}\csc x=-\csc x\cot x\) \(\{ x|x\neq k\pi, k\hbox{ an integer}\} \)
\(\dfrac{d}{dx}\sec x=\sec x\tan x\) \(\left\{ x|x\neq \dfrac{2k+1}{2}\pi, k\hbox{ an integer}\right\} \)

189

Evaluating the Second Derivative of a Trigonometric Function

Find \(f'' \left( \dfrac{\pi }{4}\right) \) if \(f( x) =\sec x\).

NOTE

If the trigonometric function begins with the letter \(c\), that is, cosine, cotangent, or cosecant, then its derivative has a minus sign.

Solution If \(f( x) =\sec x\), then \(f^\prime (x) =\sec x\tan x\) and \[ \begin{eqnarray*} f'' ( x) &=&\dfrac{d}{dx}( \sec x\tan x) \underset{\underset{\color{#0066A7}{\hbox{Use the Product Rule.}}}{{{\color{#0066A7}\uparrow }}}}= \sec x\!\left( \dfrac{d}{dx}\tan x\right) +\left( \dfrac{d}{dx}\sec x\right) \tan x \\ \\ &=&\sec x\cdot \sec ^{2}x+( \sec x\tan x) \tan x=\sec ^{3}x+\sec x\tan ^{2}x \end{eqnarray*} \]

Now, \[ \begin{eqnarray*} f'' \left( \dfrac{\pi }{4}\right) =\sec ^{3}\!\left( \dfrac{\pi }{4} \right) +\sec\! \left( \dfrac{\pi }{4}\right) \tan ^{2}\!\left( \dfrac{\pi }{4} \right) \underset{\underset{\color{#0066A7}{\hbox{sec} \dfrac{\pi }{4}=\sqrt{2};\hbox{ tan}\dfrac{\pi }{4}=1}} {{{\color{#0066A7}\uparrow }}}}= ( \sqrt{2}) ^{3}+\sqrt{2}\cdot 1^{2}=2 \sqrt{2}+\sqrt{2}=3\sqrt{2} \\ \end{eqnarray*} \]

NOW WORK

Problem 45.

Simple harmonic motion is a repetitive motion that can be modeled by a trigonometric function. A swinging pendulum and an oscillating spring are examples of simple harmonic motion.

Analyzing Simple Harmonic Motion

An object hangs on a spring, making it 2 m long in its equilibrium position. See Figure 28. If the object is pulled down 1 m and released, it will oscillate up and down. The length \(l\) of the spring after \( t\) seconds is modeled by the function \(l( t) =2+\cos t\).

Figure 28
  1. How does the length of the spring vary?
  2. Find the velocity of the object.
  3. At what position is the speed of the object a maximum?
  4. Find the acceleration of the object.
  5. At what position is the acceleration equal to 0?

Solution

  1. Since \(l(t) =2+\cos t\) and \(-1 ≤ \cos t ≤ 1,\) the length of the spring oscillates between 1 m and 3 m.
  2. The velocity \(v\) of the object is \[ v=l' ( t) =\dfrac{d}{dt}\left( 2+\cos t\right) =-\sin t \]
  3. Speed is the magnitude of velocity. Since \(v=-\sin t,\) the speed of the object is \(\left\vert v\right\vert =\left\vert -\sin t\right\vert =\left\vert \sin t\right\vert\). Since \(-1 ≤ \sin t ≤ 1,\) the object moves the fastest when \(\left\vert v\right\vert =\left\vert \sin t\right\vert =1.\) This occurs when \(\sin t=\pm 1\) or, equivalently, when \( \cos t=0.\) So, the speed is a maximum when \(l( t) =2,\) that is, when the spring is at the equilibrium position.
  4. The acceleration \(a\) of the object is given by \[ a=l'' ( t) =\dfrac{d}{dt}l' (t)=\dfrac{d}{dt} ( -\sin t) =-\cos t \]
  5. Since \(a=-\cos t,\) the acceleration is zero when \(\cos t=0.\) This occurs when \(l( t) =2,\) that is, when the spring is at the equilibrium position. At this time, the speed is maximum.

190

Figure 29 shows the graphs of the length of the spring \(y=l( t)\), the velocity \(y=v( t)\), and the acceleration \(y=a( t)\).

Figure 29

NOW WORK

Problem 65.