OBJECTIVES

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

1. Differentiate trigonometric functions (p. 185)

NEED TO REVIEW?

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

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}

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

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}

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.

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.

Evaluating the Second Derivative of a Trigonometric Function

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

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.

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.

NOW WORK

Problem 65.