Skip to main content

Workbook for Stewart’s "Calculus"

Cory Wilson

Section 2.4 Derivatives of Trigonometric Functions

Subsection 2.4.1 Before Class

https://mymedia.ou.edu/media/2.4-1/1_rapvl442
Figure 16. Pre-Class Video 1
https://mymedia.ou.edu/media/2.4-2/1_xt9ykj40
Figure 17. Pre-Class Video 2

Subsubsection 2.4.1.1 Special Trig Limits

There are two limits of trigonometric functions that we will take advantage of in Section 2.4.
Two Important Limits.
\begin{equation*} \ds \lim_{\theta\to 0} \dfrac{\sin \theta}{\theta} = 1\qquad\text{ and }\qquad \ds \lim_{\theta\to 0} \dfrac{\cos \theta - 1}{\theta} = 0 \end{equation*}
The proof refers to the following picture:
This is a circle of radius 1 with a line drawn from the origin to the point \(A = (1,0)\text{,}\) another drawn from the origin to \(D=(1,1)\) passing through the point on the circle \(B\text{.}\) A line segment is drawn from \(D\) to \(A\text{,}\) and another is drawn from \(B\) to some point \(E\) on the line segment \(AD\text{.}\) A final line segment is drawn from \(B\) to the line \(OA\text{,}\) whose intersection is point \(C\text{.}\)
First note that since we are on a unit circle, \(\sin\theta = |BC|\) which must be smaller than \(|AB|\text{.}\) In turn, \(|AB| \lt \text{Arc } AB = \theta\text{.}\) Rearranging, we see that
\begin{equation*} \sin\theta \lt \theta \iff \dfrac{\sin\theta}{\theta} \lt 1 \end{equation*}
Now, we know that
\begin{align*} \theta = \text{Arc }AB &\lt |AE| + |EB|\\ &\lt |AE| + |ED|\\ &= |AD|\\ &= \tan\theta \end{align*}
Thus, \(\theta \lt \tan \theta\)
If \(\theta \lt \tan \theta\text{,}\) then \(\theta \lt \dfrac{\sin\theta}{\cos\theta}\iff \cos\theta \lt \dfrac{\sin\theta}{\theta}\text{.}\) Combining with our first result, we have that
\begin{equation*} \cos\theta \lt \dfrac{\sin\theta}{\theta} \lt 1 \end{equation*}
Now apply the Squeeze Theorem. \(\ds \lim_{\theta\to 0} \cos\theta = \lim_{\theta\to 0} 1 = 1\text{,}\) so the Squeeze Theorem gives that
\begin{equation*} \ds \lim_{\theta\to 0} \dfrac{\sin\theta}{\theta} = 1 \end{equation*}
The second limit is a matter of computation:
\begin{align*} \ds lim_{\theta\to 0} \dfrac{\cos\theta -1}{\theta} &= lim_{\theta\to 0} \dfrac{\cos\theta -1}{\theta} \cdot \dfrac{\cos\theta + 1}{\cos\theta + 1}\\ &= \ds \lim_{\theta\to 0} \dfrac{\cos^2\theta -1}{\theta (\cos\theta + 1)}\\ &= \ds \lim_{\theta \to 0} \dfrac{\sin^2\theta}{\theta(\cos\theta +1)}\\ &= \ds \lim_{\theta \to 0} \sin\theta \cdot \dfrac{\sin\theta}{\theta} \cdot \dfrac{1}{\cos\theta + 1}\\ & = 0\cdot 1\cdot \dfrac{1}{2}\\ & = 0 \end{align*}

Subsubsection 2.4.1.2 Derivative of Sine

Example 2.4.1.
Use the limits above to show that \(\dfrac{d}{dx}\left[\sin x\right] = \cos x\text{.}\)
Solution.
We’ll use the definition of the derivative and the sum identity for sine: \(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \sin\beta\cos\alpha\text{.}\)
\begin{align*} \ds \lim_{h\to 0} \dfrac{\sin(\theta + h) - \sin\theta}{h} & = \lim_{h\to 0} \dfrac{\sin\theta \cos h + \sin h \cos\theta - \sin\theta}{h}\\ &= \ds \dfrac{\sin\theta (\cos h -1) + \sin h \cos\theta}{h} \\ &= \ds \sin\theta\cdot \lim_{h\to 0} \dfrac{\cos h -1}{h} + \cos\theta \cdot \lim_{h \to 0} \dfrac{\sin h}{h}\\ &= \cos\theta \end{align*}
Example 2.4.2.
Find the derivative of \(f(x) = \cos x\text{.}\)
Solution.
We’ll use the same approach as we did for sine, but will use the sum identity for cosine: \(\cos (\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\)
\begin{align*} \ds \lim_{h\to 0} \dfrac{\cos(\theta + h) - \cos\theta}{h} & = \lim_{h\to 0} \dfrac{\cos\theta \cos h - \sin h \sin\theta - \cos\theta}{h}\\ &= \ds \dfrac{\cos\theta (\cos h -1) - \sin h \sin\theta}{h} \\ &= \ds \cos\theta\cdot \lim_{h\to 0} \dfrac{\cos h -1}{h} - \sin\theta \cdot \lim_{h \to 0} \dfrac{\sin h}{h}\\ &=-\sin\theta \end{align*}

Subsubsection 2.4.1.3 Other Trig Derivatives

Example 2.4.3.
Find the derivatives of the following functions using the rules from Section 2.3 and the definition of the remaining four functions:
  1. \(\displaystyle \tan x\)
  2. \(\displaystyle \cot x\)
  3. \(\displaystyle \sec x\)
  4. \(\displaystyle \csc x\)
Solution.
  1. \(\displaystyle \sec^2x\)
  2. \(\displaystyle -\csc^2x\)
  3. \(\displaystyle \sec x\tan x\)
  4. \(\displaystyle -\csc x\cot x\)

Subsection 2.4.2 Pre-Class Activities

Example 2.4.4.

Find \(f'(x)\text{,}\) if \(f(x) = x^3\cos x\text{.}\)
Solution.
\(f'(x) = 3x^2\cos x - x^3\sin x\)

Example 2.4.5.

Find the equation of the tangent line to the curve \(y = 2\sin x - \cos x\) at the point \((0,-1)\text{.}\)
Solution.
\(y=2x-1\)

Example 2.4.6.

Find the equation of the tangent line and normal line to the curve \(y=(1+x)\sin x\) at the point \((0,0)\text{.}\)
Solution.
The tangent line is \(y= x\) and the normal line is \(y=-x\)

Example 2.4.7.

Find the derivative of the funciton \(k(x) = \dfrac{1+\tan x}{\sin x}\)
Solution.
\(k'(x) = \dfrac{\sin x\sec^2x - \cos x(1+\tan x)}{\sin^2x}\)

Subsection 2.4.3 In Class

Subsubsection 2.4.3.1 Trig Derivatives

Derivatives of Trigonometric Functions
\(\dfrac{d}{dx}[\sin x]\) \(\cos x\) \(\dfrac{d}{dx}[\csc x]\) \(-\csc x\cot x\)
\(\dfrac{d}{dx}[\cos x]\) \(-\sin x\) \(\dfrac{d}{dx}[\sec x]\) \(\sec x \tan x\)
\(\dfrac{d}{dx}[\tan x]\) \(\sec^2x\) \(\dfrac{d}{dx}[\cot x]\) \(-\csc^2x\)
Example 2.4.8.
For what values of \(x\) does \(f(x) = \dfrac{\cos x}{1-\sin x}\) have a horizontal tangent?
Solution.
There are no horizontal tangents
Example 2.4.9.
An object at the end of a vertical spring is stretched 5 cm beyond its resting position, and released at time \(t = 0\text{.}\) Its position at time \(t\) is given by \(s(t) = 5\sin t\text{.}\)
  1. Find the velocity and acceleration at time \(t\)
  2. At what time(s) is the velocity of the object 0? What about the acceleration?
Example 2.4.10.
Differentiate \(y = u(a\cos u + b\cot u)\text{,}\) where \(a,b\) are constants.
Solution.
\(y' = a\cos u + b\cot u + u(-a\sin u - b\csc^2u)\)
Example 2.4.11.
Differentiate \(y = x^2\sin x\tan x\)
Solution.
\(y' = 2x\sin x\tan x + x^2\cos x\tan x + x^2\sin x\sec^2x\)
Example 2.4.12.
Find the derivative of \(y = \dfrac{t\sin t}{1+t}\)
Solution.
\(y' = \dfrac{(\sin t + t\cos t)(1+t) - t\sin t}{(1+t)^2}\)
Example 2.4.13.
Find the 101st derivative of \(f(x) = 2\cos x\text{.}\)
Solution.
\(\dfrac{d^{101}f}{dx^{101}} = -2\sin x\)
Example 2.4.14.
Find the 99th derivative of \(f(x) = \sin x\)
Solution.
\(f^{99}(x) = -\cos x\)
Example 2.4.15.
Find the equation of the tangent line to the curve \(y = x+\tan x\) at the point \((\pi,\pi)\text{.}\)
Solution.
\(y=\pi\)

Subsection 2.4.4 After Class Activities

Example 2.4.16.

Suppose \(f(\pi/3) = 4\text{,}\) and \(f'(\pi/3) = -2\text{.}\) Let \(g(x) = f(x)\sin x\) and \(h(x) = \dfrac{\cos x}{f(x)}\text{.}\) Find \(g'(\pi/3)\) and \(h'(\pi/3)\text{.}\)
Solution.
\(g'\lrpar{{\pi}{3}} = 2-\sqrt{3}\) and \(h'\lrpar{\dfrac{\pi}{3}} = \dfrac{1-2\sqrt{3}}{16}\)

Example 2.4.17.

Find \(\dfrac{d^{35}}{dx^{35}} (x\sin x)\)
Solution.
\(\sin x - x\cos x\)

Example 2.4.18.

Find constants \(A\) and \(B\) so that, for \(y = A\sin x + B\cos x\text{,}\) the following holds:
\begin{equation*} y'' + y' - 2y = \sin x \end{equation*}
Hint: If \(y = A\sin x + B\cos x\text{,}\) then \(y' = A\cos x - B\sin x\text{,}\) and \(y'' = -A\sin x - B\cos x\text{.}\)
Solution.
\(A = -\dfrac{3}{10},B = -\dfrac{1}{10}\)

Example 2.4.19.

Find the points on the curve \(y = \dfrac{\cos x}{2 + \sin x}\) where the graph has a horizontal tangent.
Solution.
\(x = \dfrac{\pi}{6} + 2\pi k,\dfrac{5\pi}{6} + 2\pi k\text{,}\) for \(k\in \Z\)
<Prev^TopNext>