Derivative Rules for Trigonometric Functions

Section 4.5 Derivative Rules for Trigonometric Functions

We next look at the derivative of the sine function. In order to prove the derivative formula for sine, we recall two limit computations from earlier:

\begin{equation*} \lim_{x\to 0}\frac{\sin x}{x}=1\qquad\mbox{ and } \qquad\lim_{x\to 0}\frac{\cos x -1}{x}=0\text{,} \end{equation*}

and the double angle formula

\begin{equation*} \sin\left(A+B\right)=\sin A\cos B+\sin B\cos A\text{.} \end{equation*}

Theorem 4.53. Derivative of Sine Function.

$\ds{\left(\sin x\right)'=\cos x}$

Proof.

Let $f(x)=\sin x\text{.}$ Using the definition of derivative we have:

\begin{equation*} \begin{array}{ccl} f'(x) \amp = \amp \ds{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}}\\ \\ \amp = \amp \ds{\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}}\\ \\ \amp = \amp \ds{\lim_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}}\\ \\ \amp = \amp \ds{\lim_{h\to 0}\sin x\cdot\lim_{h\to 0}\frac{\cos h-1}{h}+\lim_{h\to 0}\cos x\cdot\lim_{h\to 0}\frac{\sin h}{h}}\\ \\ \amp = \amp \ds{\sin x\cdot 0+\cos x\cdot 1}\\ \\ \amp = \amp \ds{\cos x} \end{array} \end{equation*}

since

\begin{equation*} \lim_{x\to 0}\frac{\sin x}{x}=1\quad\mbox{and}\quad \lim_{x\to 0}\frac{\cos x -1}{x}=0\text{.} \end{equation*}

Interactive Demonstration. The graph below shows the graph of $\sin x$ (in blue) and the associated tangent line (in grey) at each point on the graph. At each point, the graph of the derivative is plotted in red. By dragging the red point, investigate the derivative $(\sin x)' $ and notice that $(\sin x)' = \cos x\text{:}$

A formula for the derivative of the cosine function can be found in a similar fashion:

\begin{equation*} \frac{d}{dx}(\cos x)=-\sin x\text{.} \end{equation*}

Using the Quotient Rule we get formulas for the remaining trigonometric ratios. To summarize, here are the derivatives of the six trigonometric functions:

Theorem 4.54. Derivatives of Basic Trigonometric Functions.

\begin{equation*} \begin{array}{ll} \ds{\frac{d}{dx}\left(\sin(x)\right)} = \ds{\cos(x)} \amp \ds{\frac{d}{dx}\left(\cos(x)\right)} = \ds{-\sin(x)} \\ \ds{\frac{d}{dx}\left(\tan(x)\right)} = \ds{\sec^2(x)} \amp \ds{\frac{d}{dx}\left(\csc(x)\right)} = \ds{-\csc(x)\cot(x)}\\ \ds{\frac{d}{dx}\left(\sec(x)\right)} = \ds{\sec(x)\tan(x)} \amp \ds{\frac{d}{dx}\left(\cot(x)\right)} = \ds{-\csc^2(x)} \end{array} \end{equation*}

Example 4.55. Derivative of Product of Trigonometric Functions.

Find the derivative of $f(x)=\sin x\tan x\text{.}$

Solution

Using the Product Rule we obtain

\begin{equation*} f'(x)=\cos x\tan x+\sin x\sec^2x\text{.} \end{equation*}

Example 4.56. Chain Rule with Trigonometric Functions.

Differentiate each of the following functions:

$h(x)=(x+\sin(x^{2}))^{10}$
$k(x)=\sin(\cos^{2}x)$

Solution

We use the Chain Rule and the general Power Rule to find

\begin{equation*} h'(x) = 10(x+\sin x^{2})^{9} \frac{d}{dx}(x+\sin x^{2}) \end{equation*}

We apply the Chain Rule once more using the derivatives of trigonometric functions:

\begin{equation*} \begin{split} h'(x) \amp = 10(x+\sin x^{2})^{9} \left[1+\cos(x^{2})\frac{d}{dx}(x^{2})\right]\\ \amp = 10(x+\sin x^{2})^{9}(1+2x\cos(x^{2})) \end{split} \end{equation*}
We first rewrite $k(x)$ as $k(x)=\sin((\cos x)^{2})\text{.}$ Then

\begin{equation*} \begin{split} k'(x) \amp = \cos(\cos^{2}x)\frac{d}{dx}((\cos x)^{2}) \\ \amp = \cos(\cos^{2}x) 2\cos x \frac{d}{dx}(\cos x) \\ \amp = \cos(\cos^{2}x) 2\cos x (-\sin x) \\ \amp = -2(\sin x)(\cos x)(\cos(\cos^{2}x)) \end{split} \end{equation*}

Example 4.57. Tangent Line.

Find an equation of the tangent line to the graph of the function $\ds f(x)=\tan 2x$ at the point $\left(\frac{\pi}{8},1\right)\text{.}$

Solution

The slope of the tangent line at any point on the graph of $f$ is given by

\begin{equation*} f'(x) = 2 \sec^{2}(2x)\text{.} \end{equation*}

In particular, the slope of the tangent line at the point $\left(\frac{\pi}{8},1\right)$ is given by

\begin{equation*} f'\left(\frac{\pi}{8}\right) = 2\sec^{2}\left(\frac{\pi}{4}\right) = 2\left(\frac{2}{\sqrt{2}}\right)^{2} = 4\text{.} \end{equation*}

Therefore, a required equation is given by

\begin{equation*} y-1 = 4\left(x-\frac{\pi}{8}\right)\text{,} \end{equation*}

or,

\begin{equation*} y= 4x + \left(1-\frac{\pi}{2}\right)\text{.} \end{equation*}

Exercises for Section 4.5.

Exercise 4.5.1.

Find the derivatives of the following functions.

$\ds f(x) = \sin x\cos x$
Answer
$\ds f'(x)= \cos 2x$
Solution
$\begin{aligned}\diff{f}{x} \amp = \diff{}{x} \sin x \cos x \\ \amp = \cos x \diff{}{x} \sin x + \sin x \diff{}{x} \cos x \\ \amp = \cos x \cos x + \sin x (-\sin x) \\ \amp = \cos^2 x - \sin^2 x \end{aligned}$
$\ds f(x) = \cot x$
Answer
$\ds f'(x)= -\csc^{2}x$
Solution
$\begin{aligned}f'(x) \amp = \diff{}{x} \cot x \\ \amp = \diff{}{x} \frac{\cos x }{\sin x} \\ \amp = \frac{\sin x (\cos x)' - \cos x (\sin x)'}{\sin^2 x} \\ \amp = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\\ \amp = \frac{-1}{\sin^2 x} \\ \amp = - \csc^2 x \end{aligned}$
$\ds f(x) = \csc x-x\tan x$
Answer
$\ds f'(x)= -\tan x - x\sec^{2}x - \cot x\csc x$
Solution
$\begin{aligned}\diff{f}{x} \amp = \diff{}{x}\left(\csc(x)-x\tan(x)\right)\amp \\\amp = \diff{}{x}\left(\frac{1}{\sin(x)}\right)-\left(\tan(x)-x\diff{}{x}\left(\frac{\sin(x)}{\cos(x)}\right)\right)\amp \\\amp = -\frac{\cos(x)}{\sin^{2}(x)} - \tan(x) - \frac{\cos^{2}(x)+\sin^{2}(x)}{\cos^{2}(x)} \amp \\\amp = -\frac{\cos(x)}{\sin^{2}(x)} - \tan(x) - \frac{1}{\cos^{2}(x)} \amp\\ \amp = -\cot(x)\csc(x) - \tan(x) - x\sec^{2}(x)\amp \end{aligned}$
$\ds h(x) = 2\cos \pi x$
Answer
$\ds h'(x)= -2\pi \sin \pi x$
Solution
$\begin{aligned}h'(x) \amp = \diff{}{x} 2 \cos(\pi x) \\ \amp = -2\sin (\pi x) \diff{}{x} (\pi x) \\ \amp = - 2 \pi \sin (\pi x) \end{aligned}$
$\ds f(t) = \sin(t^{2}+1)$
Answer
$\ds f'(t)= 2t\cos(t^{2}+1)$
Solution
$\begin{aligned}f'(t) \amp = \diff{}{t} \sin (t^2+1) \\ \amp = \cos (t^2+1) \diff{}{t} (t^2+1)\\ \amp = 2t\cos(t^2+1) \end{aligned}$
$\ds f(x) = \tan 2x^{2}$
Answer
$\ds f'(x)= 4x\sec^{2}2x^{2}$
Solution
$\begin{aligned}f'(x) \amp = \diff{}{x} \tan(2x^2) \\ \amp = \sec^2(2x^2)\diff{}{x} (2x^2) \\ \amp = 4x \sec^2(2x^2) \end{aligned}$
$\ds g(x) = x \sin x$
Answer
$\ds g'(x)= x\cos x + \sin x$
Solution
$\begin{aligned}g'(x) \amp = \diff{}{x} x\sin x \\ \amp = \sin x \diff{}{x} (x) + x \diff{}{x} \sin x \\ \amp = \sin x + x \cos x \end{aligned}$
$\ds f(x) = 2\sin 3x + 3\cos 2x$
Answer
$\ds f'(x)= 6(\cos 3x - \sin 2x)$
Solution
$\begin{aligned}f'(x) \amp = \diff{}{x} \left(2\sin 3x + 3 \cos 2x\right) \\ \amp = 2\cos(3x) \diff{}{x} (3x) - 3 \sin (2x) \diff{}{x} (2x) \\ \amp = 6\cos(3x) - 6 \sin (2x) \end{aligned}$
$\ds f(s) = 2\cot 2s + \sec 3s$
Answer
$\ds f'(s)= -4\csc^2(2s) + 3 \tan(3s)\sec(3s)$
Solution
$\begin{aligned}f'(s) \amp = \diff{}{s} \left(2\cot (2s) + \sec(3s) \right) \\ \amp = 2 (-\csc^2(2s)) \diff{}{s} (2s) + \tan(3s)\sec(3s)\diff{}{s}(3s)\\ \amp = -4\csc^2(2s) + 3 \tan(3s)\sec(3s) \end{aligned}$
$\ds f(x) = x^{2}\cos 2x$
Answer
$f'(x) =2x\cos x -x^2\sin x $
Solution
$\begin{aligned}f'(x) \amp = \diff{}{x} x^2\cos(x) \\ \amp = \cos x \diff{}{x} (x^2) + x^2 \diff{}{x} \cos(x) \\ \amp = 2x\cos x -x^2\sin x \end{aligned}$
$\ds h(s) = \sin\sqrt{s^{2}-1}$
Answer
$\ds h'(s)= \frac{s\cos(\sqrt{s^{2}-1})}{\sqrt{s^{2}-1}}$
Solution
$\begin{aligned}\diff{h}{s} \amp = \diff{}{s}\left(\sin(\sqrt{s^{2}-1})\right)\amp \\ \amp = \cos(\sqrt{s^{2}-1}) \cdot \diff{}{s}\left(\sqrt{s^{2}-1}\right) \amp\\ \amp = \frac{\cos(\sqrt{s^{2}-1})}{2\sqrt{s^{2}-1}} \cdot 2s \amp \\ \amp = \frac{s\cos(\sqrt{s^{2}-1})}{\sqrt{s^{2}-1}} \amp \end{aligned}$
$\ds f(x) = x\cos\frac{1}{x}$
Answer
$f'(x)= \frac{x\sin (1/x)}{x^2} + \cos(1/x)$
Solution
$\begin{aligned}f'(x) \amp = \diff{}{x} x \cos(1/x) \\ \amp = x \diff{}{x} \cos(1/x) + \cos(1/x)\diff{}{x} (x) \\ \amp = -x\sin(1/x) \diff{}{x}\frac{1}{x} + \cos(1/x)\\ \amp = -x \sin(1/x) \frac{-1}{x^2} + \cos(1/x) \\ \amp = \frac{x\sin (1/x)}{x^2} + \cos(1/x) \end{aligned}$
$\ds f(x) = \dfrac{x-\sin x}{1+\cos x}$
Answer
$\ds f'(x)= \frac{x\sin (1/x)}{x^2} + \cos(1/x$
Solution
$\begin{aligned}f'(x) \amp = \diff{}{x} x \cos(1/x) \\ \amp = x \diff{}{x} \cos(1/x) + \cos(1/x)\diff{}{x} (x) \\ \amp = -x\sin(1/x) \diff{}{x}\frac{1}{x} + \cos(1/x)\\ \amp = -x \sin(1/x) \frac{-1}{x^2} + \cos(1/x) \\ \amp = \frac{x\sin (1/x)}{x^2} + \cos(1/x) \end{aligned}$
$\ds g(t) = \sqrt{\tan t}$
Answer
$\ds g'(t)= \dfrac{\sec^{2}t}{2\sqrt{\tan t}}$
Solution
$\begin{aligned}g'(t) \amp = \diff{}{t} \sqrt{\tan t} \\ \amp = \frac{1}{2\sqrt{\tan t}} \diff{}{t} \tan t\\ \amp = \frac{1}{2\sqrt{\tan t}} \sec^2 t \end{aligned}$

Exercise 4.5.2.

Find the points on the curve $\ds y= x+ 2\cos x$ that have a horizontal tangent line.

Answer

$\ds \pi/6+2n\pi\text{,}$ $5\pi/6+2n\pi\text{,}$ any integer $n$

Solution

Let $f(x) = x+2\cos x\text{.}$ Then

\begin{equation*} f'(x) = \diff{}{x} (x+2\cos x) = 1-2 \sin x\text{.} \end{equation*}

The tangent line to $f$ at some point $a$ is horizontal if $f'(a) = 0\text{.}$ Therefore, we solve

\begin{equation*} \begin{split} f'(a) \amp = 0 \\ 1- 2 \sin a = 0 \\ \sin a = \frac{1}{2} \end{split} \end{equation*}

Consider the following right triangle:

And so in the interval $\left[0,\frac{\pi}{2}\right]\text{,}$ we have $a= \frac{\pi}{6}\text{.}$ And so from $\left[0,2\pi\right]\text{,}$ we have two solutions: $a = \frac{\pi}{6}$ and $a=\frac{5\pi}{6}\text{.}$

Therefore, at all points $x = \frac{\pi}{6} + 2 k \pi$ for any integer $k\text{,}$ the graph of $f(x)$ has a horizontal tangent line.

Exercise 4.5.3.

The revenue of McMenamy's Fish Shanty located at a popular summer resort is approximately

\begin{equation*} R(t)=2\left(5-4\cos\frac{\pi}{6}t\right) \ \ \ 0\leq t \leq 12 \end{equation*}

during the $t$-th week ($t=1$ corresponds to the first week of June), where $R$ is measured in thousands of dollars. On what week is the marginal revenue zero?

Answer

$t=0,6$ and $12$

Solution

Given

\begin{equation*} R(t) = 2\left(5-4\cos \frac{\pi}{6}t\right)\text{,} \end{equation*}

for $t \in \left[0,12\right]\text{,}$ we find the marginal revenue by taking the derivative with respect to $t\text{.}$

\begin{equation*} R'(t) = 2\diff{ }{t}\left(5-4\cos \frac{\pi}{6}t\right) = -8\left(\diff{ }{t}\cos \frac{\pi}{6}t\right)\text{.} \end{equation*}

We now use the fact that $\diff{ }{t} \cos t = -\sin t$ and apply the chain rule to find

\begin{equation*} R'(t)=\frac{4}{3}\pi \sin \frac{\pi}{6}t\text{.} \end{equation*}

Therefore, $R'(t) = 0$ when $\sin \frac{\pi}{6}t = 0 \implies \frac{\pi}{6}t = n\pi\text{,}$ or $t = 6n$ where $n=0,1,2$ (due to our restrictions on $t$). The marginal revenue is thus zero on the $0$th, $6$th and $12$th weeks.

Exercise 4.5.4.

The revenue in thousands of dollars received from the sale of electronic fans is seasonal, with maximum revenue in the summer. Let the revenue received from the sales of fans be approximated by

\begin{equation*} R(x)=-100\cos(2\pi x)+120 \end{equation*}

where $x$ is time in years, measured from January $1\text{.}$ Calculate the marginal revenue for September $1\text{.}$ (Take that one month represents $1/12$ of the years).

Answer

-$$628.32$ thousand dollars per year.

Solution

We first differentiate:

\begin{equation*} R'(x) = \diff{}{x} \left(-100\cos(2\pi x) + 120\right) = 200\pi \sin(2\pi x) \end{equation*}

Since $x$ is the time in years, the marginal revenue for September 1 is found by taking $x = 9/12 = 3/4\text{.}$ Thus,

\begin{equation*} R'(3/4) = 200 \pi \sin (3 \pi /2) = -200 \pi\text{.} \end{equation*}

Therefore, on Septermber 1, the revenue was decreasing by approximately $628.32 per year.

Exercise 4.5.5.

Sales of large kitchen appliances such as ovens and fridges are usually subject to seasonal fluctuations. Everything Kitchen's sales of fridge models from the beginning of $2001$ to the end of $2002$ can be approximated by

\begin{equation*} S(x)=\frac{1}{10}\sin\left(\frac{\pi}{2}(x+1)\right)+\frac{1}{2} \end{equation*}

where $x$ is time in quarters, $x=1$ corresponds to the end of the first quarter of $2001\text{,}$ and $S$ is measured in millions of dollars. Find the rate of change of sales for the end of the third quarter of $2002\text{.}$

Answer

Rate of change of sales are increasing by $$157,079.63$ per quarter.

Solution

We differentiate:

\begin{equation*} \begin{split} S'(x) \amp = \frac{1}{10} \diff{}{x} \sin \left(\frac{\pi}{2}(x+1)\right) \\ \amp = \frac{1}{10} \cos\left(\frac{\pi}{2}(x+1)\right) \diff{}{x} \left(\frac{\pi}{2}(x+1)\right) \\ \amp = \frac{\pi}{20} \cos\left(\frac{\pi}{2}(x+1)\right) \end{split} \end{equation*}

Since $x$ is time in quarters and $x=1$ corresponds to the end of the first quarter of 2001, the end of the third quarter of 2002 is given by $x=7\text{.}$ Thus,

\begin{equation*} S'(7) = \frac{\pi}{20} \cos\left(\frac{8\pi}{2}\right) = \frac{\pi}{20}\text{.} \end{equation*}

This means that sales were increasing at a rate of approximately $157,079.63 per quarter.