Finding the Derivatives of Trig Functions

Learning Outcomes

Find the derivatives of the sine and cosine function.
Find the derivatives of the standard trigonometric functions.
Calculate the higher-order derivatives of the sine and cosine.

Derivatives of the Sine and Cosine Functions

We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function $f(x),$

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

$f^{\prime}(x)=\underset{h\to 0}{\lim}\dfrac{f(x+h)-f(x)}{h}$

Consequently, for values of $h$ very close to 0, $f^{\prime}(x)\approx \frac{f(x+h)-f(x)}{h}$ . We see that by using $h=0.01$ ,

\frac{d}{d x} (\sin x) \approx \frac{\sin (x + 0.01) - \sin x}{0.01}

$\frac{d}{dx}(\sin x)\approx \dfrac{\sin(x+0.01)-\sin x}{0.01}$

By setting $D(x)=\frac{\sin(x+0.01)-\sin x}{0.01}$ and using a graphing utility, we can get a graph of an approximation to the derivative of $\sin x$ (Figure 1).

The function D(x) = (sin(x + 0.01) − sin x)/0.01 is graphed. It looks a lot like a cosine curve.

Figure 1. The graph of the function $D(x)$ looks a lot like a cosine curve.

Upon inspection, the graph of $D(x)$ appears to be very close to the graph of the cosine function. Indeed, we will show that

\frac{d}{d x} (\sin x) = \cos x

$\frac{d}{dx}(\sin x)= \cos x$

If we were to follow the same steps to approximate the derivative of the cosine function, we would find that

\frac{d}{d x} (\cos x) = - \sin x

$\frac{d}{dx}(\cos x)=−\sin x$

The Derivatives of $\sin x$ and $\cos x$

The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.

\frac{d}{d x} (\sin x) = \cos x

$\frac{d}{dx}(\sin x)= \cos x$

\frac{d}{d x} (\cos x) = - \sin x

$\frac{d}{dx}(\cos x)=−\sin x$

Proof

Because the proofs for $\frac{d}{dx}(\sin x)= \cos x$ and $\frac{d}{dx}(\cos x)=−\sin x$ use similar techniques, we provide only the proof for $\frac{d}{dx}(\sin x)= \cos x$ . Before beginning, recall two important trigonometric limits we learned in Module 2: Limits.

lim_{h \to 0} \frac{\sin h}{h} = 1

$\underset{h\to 0}{\lim}\frac{\sin h}{h}=1$ and

lim_{h \to 0} \frac{\cos h - 1}{h} = 0

$\underset{h\to 0}{\lim}\frac{\cos h-1}{h}=0$

The graphs of $y=\frac{(\sin h)}{h}$ and $y=\frac{(\cos h-1)}{h}$ are shown in Figure 2.

The function y = (sin h)/h and y = (cos h – 1)/h are graphed. They both have discontinuities on the y-axis.

Figure 2. These graphs show two important limits needed to establish the derivative formulas for the sine and cosine functions.

We also recall the following trigonometric identity for the sine of the sum of two angles:

\sin (x + h) = \sin x \cos h + \cos x \sin h

$\sin(x+h)= \sin x \cos h+ \cos x \sin h$

Now that we have gathered all the necessary equations and identities, we proceed with the proof.

\begin{array}{lllll} \frac{d}{d x} \sin x & = lim_{h \to 0} \frac{\sin (x + h) - \sin x}{h} & Apply the definition of the derivative. \\ = lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} & Use trig identity for the sine of the sum of two angles. \\ = lim_{h \to 0} (\frac{\sin x \cos h - \sin x}{h} + \frac{\cos x \sin h}{h}) & Regroup. \\ = lim_{h \to 0} (\sin x (\frac{\cos h - 1}{h}) + \cos x (\frac{\sin h}{h})) & Factor out \sin x and \cos x . \\ = \sin x \cdot 0 + \cos x \cdot 1 & Apply trig limit formulas. \\ = \cos x & Simplify. \end{array}

$\begin{array}{lllll}\frac{d}{dx} \sin x & =\underset{h\to 0}{\lim}\frac{\sin(x+h)-\sin x}{h} & & & \text{Apply the definition of the derivative.} \\ & =\underset{h\to 0}{\lim}\frac{\sin x \cos h+ \cos x \sin h- \sin x}{h} & & & \text{Use trig identity for the sine of the sum of two angles.} \\ & =\underset{h\to 0}{\lim}\left(\frac{\sin x \cos h-\sin x}{h}+\frac{\cos x \sin h}{h}\right) & & & \text{Regroup.} \\ & =\underset{h\to 0}{\lim}\left(\sin x\left(\frac{\cos h-1}{h}\right)+ \cos x\left(\frac{\sin h}{h}\right)\right) & & & \text{Factor out} \, \sin x \, \text{and} \, \cos x. \\ & = \sin x\cdot{0}+ \cos x\cdot{1} & & & \text{Apply trig limit formulas.} \\ & = \cos x & & & \text{Simplify.} \end{array}$

$_\blacksquare$

Figure 3 shows the relationship between the graph of $f(x)= \sin x$ and its derivative $f^{\prime}(x)= \cos x$ . Notice that at the points where $f(x)= \sin x$ has a horizontal tangent, its derivative $f^{\prime}(x)= \cos x$ takes on the value zero. We also see that where $f(x)= \sin x$ is increasing, $f^{\prime}(x)= \cos x>0$ and where $f(x)= \sin x$ is decreasing, $f^{\prime}(x)= \cos x<0$ .

The functions f(x) = sin x and f’(x) = cos x are graphed. It is apparent that when f(x) has a maximum or a minimum that f’(x) = 0.

Figure 3. Where $f(x)$ has a maximum or a minimum, $f^{\prime}(x)=0$ . That is, $f^{\prime}(x)=0$ where $f(x)$ has a horizontal tangent. These points are noted with dots on the graphs.

Example: Differentiating a Function Containing $\sin x$

Find the derivative of $f(x)=5x^3 \sin x$ .

Show Solution

Using the product rule, we have

\begin{array}{ll} f^{'} (x) & = \frac{d}{d x} (5 x^{3}) \cdot \sin x + \frac{d}{d x} (\sin x) \cdot 5 x^{3} \\ = 15 x^{2} \cdot \sin x + \cos x \cdot 5 x^{3} \end{array}

$\begin{array}{ll}f^{\prime}(x) & =\frac{d}{dx}(5x^3)\cdot \sin x+\frac{d}{dx}(\sin x)\cdot 5x^3 \\ & =15x^2\cdot \sin x+ \cos x\cdot 5x^3\end{array}$

After simplifying, we obtain

f^{'} (x) = 15 x^{2} \sin x + 5 x^{3} \cos x

$f^{\prime}(x)=15x^2 \sin x+5x^3 \cos x$ .

Try It

Find the derivative of $f(x)= \sin x \cos x.$

Hint

Show Solution

$f^{\prime}(x)=\cos^2 x-\sin^2 x$

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

Try It

Example: Finding the Derivative of a Function Containing $\cos x$

Find the derivative of $g(x)=\dfrac{\cos x}{4x^2}$ .

Show Solution

By applying the quotient rule, we have

g^{'} (x) = \frac{(- \sin x) 4 x^{2} - 8 x (\cos x)}{(4 x^{2})^{2}}

$g^{\prime}(x)=\frac{(−\sin x)4x^2-8x(\cos x)}{(4x^2)^2}$ .

Simplifying, we obtain

\begin{array}{ll} g^{'} (x) & = \frac{- 4 x^{2} \sin x - 8 x \cos x}{16 x^{4}} \\ = \frac{- x \sin x - 2 \cos x}{4 x^{3}} \end{array}

$\begin{array}{ll}g^{\prime}(x) & =\frac{-4x^2 \sin x-8x \cos x}{16x^4} \\ & =\frac{−x \sin x-2 \cos x}{4x^3} \end{array}$

Try It

Find the derivative of $f(x)=\dfrac{x}{\cos x}$ .

Hint

Show Solution

$\frac{\cos x+x \sin x}{\cos^2 x}$

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

Example: An Application to Velocity

A particle moves along a coordinate axis in such a way that its position at time $t$ is given by $s(t)=2 \sin t-t$ for $0\le t\le 2\pi$ . At what times is the particle at rest?

Show Solution

To determine when the particle is at rest, set $s^{\prime}(t)=v(t)=0$ . Begin by finding $s^{\prime}(t)$ . We obtain

s^{'} (t) = 2 \cos t - 1

$s^{\prime}(t)=2 \cos t-1$ ,

so we must solve

2 \cos t - 1 = 0

$2 \cos t-1=0$ for

0 \leq t \leq 2 π

$0\le t\le 2\pi$ .

The solutions to this equation are $t=\frac{\pi}{3}$ and $t=\frac{5\pi}{3}$ . Thus the particle is at rest at times $t=\frac{\pi}{3}$ and $t=\frac{5\pi}{3}$ .

Try It

A particle moves along a coordinate axis. Its position at time $t$ is given by $s(t)=\sqrt{3}t+2 \cos t$ for $0\le t\le 2\pi$ . At what times is the particle at rest?

Show Solution

$t=\frac{\pi}{3}, \, t=\frac{2\pi}{3}$

Hint

Use the previous example as a guide.

Derivatives of Other Trigonometric Functions

Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives.

Example: The Derivative of the Tangent Function

Find the derivative of $f(x)= \tan x$ .

Show Solution

Start by expressing $\tan x$ as the quotient of $\sin x$ and $\cos x$ :

f (x) = \tan x = \frac{\sin x}{\cos x}

$f(x)= \tan x=\dfrac{\sin x}{\cos x}$

Now apply the quotient rule to obtain

f^{'} (x) = \frac{\cos x \cos x - (- \sin x) \sin x}{(\cos x)^{2}}

$f^{\prime}(x)=\dfrac{\cos x \cos x-(−\sin x)\sin x}{(\cos x)^2}$

Simplifying, we obtain

f^{'} (x) = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}

$f^{\prime}(x)=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}$

Recognizing that $\cos^2 x+\sin^2 x=1$ , by the Pythagorean Identity, we now have

f^{'} (x) = \frac{1}{\cos^{2} x}

$f^{\prime}(x)=\dfrac{1}{\cos^2 x}$

Finally, use the identity $\sec x=\dfrac{1}{\cos x}$ to obtain

f^{'} (x) = \sec^{2} x

$f^{\prime}(x)=\sec^2 x$

Try It

Find the derivative of $f(x)= \cot x$ .

Hint

Rewrite $\cot x$ as $\frac{\cos x}{\sin x}$ and use the quotient rule.

Show Solution

$f^{\prime}(x)=−\csc^2 x$

The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem.

Derivatives of $\tan x, \, \cot x, \, \sec x$ , and $\csc x$

The derivatives of the remaining trigonometric functions are as follows:

\frac{d}{d x} (\tan x) = \sec^{2} x

$\frac{d}{dx}(\tan x)=\sec^2 x$

\frac{d}{d x} (\cot x) = - \csc^{2} x

$\frac{d}{dx}(\cot x)=−\csc^2 x$

\frac{d}{d x} (\sec x) = \sec x \tan x

$\frac{d}{dx}(\sec x)= \sec x \tan x$

\frac{d}{d x} (\csc x) = - \csc x \cot x

$\frac{d}{dx}(\csc x)=−\csc x \cot x$

As you navigate problems involving derivatives of trigonometric functions, don’t forget our handy table of trigonometric function values of common angles:

Recall: Trigonometric function values of common angles

Angle	$0$	$\frac{\pi }{6},\text{ or }{30}^{\circ}$	$\frac{\pi }{4},\text{ or } {45}^{\circ }$	$\frac{\pi }{3},\text{ or }{60}^{\circ }$	$\frac{\pi }{2},\text{ or }{90}^{\circ }$
Cosine	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
Sine	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
Tangent	0	$\frac{\sqrt{3}}{3}$	1	$\sqrt{3}$	Undefined
Secant	1	$\frac{2\sqrt{3}}{3}$	$\sqrt{2}$	2	Undefined
Cosecant	Undefined	2	$\sqrt{2}$	$\frac{2\sqrt{3}}{3}$	1
Cotangent	Undefined	$\sqrt{3}$	1	$\frac{\sqrt{3}}{3}$	0

Example: Finding the Equation of a Tangent Line

Find the equation of a line tangent to the graph of $f(x)= \cot x$ at $x=\dfrac{\pi}{4}$ .

Show Solution

To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute

f (\frac{π}{4}) = \cot \frac{π}{4} = 1

$f\left(\frac{\pi}{4}\right)= \cot \frac{\pi}{4}=1$ .

Thus the tangent line passes through the point $\left(\frac{\pi}{4},1\right)$ . Next, find the slope by finding the derivative of $f(x)= \cot x$ and evaluating it at $\frac{\pi}{4}$ :

f^{'} (x) = - \csc^{2} x

$f^{\prime}(x)=−\csc^2 x$ and

f^{'} (\frac{π}{4}) = - \csc^{2} (\frac{π}{4}) = - 2

$f^{\prime}\left(\frac{\pi}{4}\right)=−\csc^2 \left(\frac{\pi}{4}\right)=-2$ .

Using the point-slope equation of the line, we obtain

y - 1 = - 2 (x - \frac{π}{4})

$y-1=-2\left(x-\frac{\pi}{4}\right)$

or equivalently,

y = - 2 x + 1 + \frac{π}{2}

$y=-2x+1+\frac{\pi}{2}$ .

Example: Finding the Derivative of Trigonometric Functions

Find the derivative of $f(x)= \csc x+x \tan x.$

Show Solution

To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find

f^{'} (x) = \frac{d}{d x} (\csc x) + \frac{d}{d x} (x \tan x)

$f^{\prime}(x)=\frac{d}{dx}(\csc x)+\frac{d}{dx}(x \tan x)$ .

In the first term, $\frac{d}{dx}(\csc x)=−\csc x \cot x$ , and by applying the product rule to the second term we obtain

\frac{d}{d x} (x \tan x) = (1) (\tan x) + (\sec^{2} x) (x)

$\frac{d}{dx}(x \tan x)=(1)(\tan x)+(\sec^2 x)(x)$ .

Therefore, we have

f^{'} (x) = - \csc x \cot x + \tan x + x \sec^{2} x

$f^{\prime}(x)=−\csc x \cot x+ \tan x+x \sec^2 x$ .

Watch the following video to see the worked solution to Example: Finding the Derivative of Trigonometric Functions.

Closed Captioning and Transcript Information for Video

Try It

Find the derivative of $f(x)=2 \tan x-3 \cot x$ .

Hint

Show Solution

$f^{\prime}(x)=2 \sec^2 x+3 \csc^2 x$

Try It

Find the slope of the line tangent to the graph of $f(x)= \tan x$ at $x=\frac{\pi}{6}$ .

Hint

Evaluate the derivative at $x=\frac{\pi}{6}$ .

Show Solution

$\frac{4}{3}$

Module 3: Derivatives