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# 2.4: Derivatives of Trigonometric Functions

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Compute the derivatives of various trigonometric functions.

If the angle $h$ is measured in radians,

$\lim_{h \to 0} \frac{\sin h} {h} = 1$ and $\lim_{h \to 0} \frac{1 - \cos h} {h} = 0.$

We can use these limits to find an expression for the derivative of the six trigonometric functions $\sin x, \cos x, \tan x, \sec x, \csc x,$ and $\cot x$. We first consider the problem of differentiating $\sin x$, using the definition of the derivative.

$\frac{d} {dx} [\sin x] = \lim_{h \to 0} \frac{\sin(x + h) - \sin x} {h}$

Since

$\sin(\alpha + \beta) = \sin \alpha\ \cos \beta + \cos \alpha\ \sin \beta.$

The derivative becomes

$\frac{d} {dx} [\sin x] & = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x} {h}\\& = \lim_{h \to 0} \left [\sin x \left (\frac{\cos h - 1} {h} \right ) + \cos x \left (\frac {\sin h} {h} \right) \right]\\& = -\!\sin x \cdot \lim_{h \to 0} \left (\frac{1 - \cos h} {h} \right) + \cos x \cdot \lim_{h \to 0} \left (\frac{\sin h} {h} \right)\\& = -\!\sin x \cdot (0) + \cos x \cdot (1)\\& = \cos x.$

Therefore,

$\frac {d}{dx} [\sin x] = \cos x.$

It will be left as an exercise to prove that

$\frac {d}{dx}[\cos x] = -\!\sin x.$

The derivatives of the remaining trigonometric functions are shown in the table below.

Derivatives of Trigonometric Functions

$\frac {d}{dx}[\sin x] & = \cos x\\\frac {d}{dx}[\cos x] & = -\!\sin x\\\frac {d}{dx}[\tan x] & = {\sec^2x}\\\frac {d}{dx}[\sec x] & = {\sec x}\ {\tan x}\\\frac {d}{dx}[\csc x] & = {-\!\csc x}\ {\cot x}\\\frac {d}{dx}[\cot x] & = {-\!\csc^2 x}$

Keep in mind that for all the derivative formulas for the trigonometric functions, the argument $x$ is measured in radians.

Example 1:

Show that $\frac {d}{dx} [\tan x] = \sec^2 x.$

Solution:

It is possible to prove this relation by the definition of the derivative. However, we use a simpler method.

Since

$\tan x =\frac{\sin x}{\cos x},$

then

$\frac {d}{dx}[\tan x] &= \frac {d}{dx}\left [\frac{\sin x}{\cos x}\right]\\\text{Using~the~quotient~rule,}\\&= \frac{(\cos x)(\cos x) - (\sin x)(-\!\sin x)}{\cos^2 x}\\&= \frac {\cos^2 x + \sin^2 x}{\cos^2 x}\\&= \frac {1}{\cos^2 x}\\&= {\sec^2 x}$

Example 2:

Find $f'(x) \mathrm{~if~} f(x) = x^{2} \cos x + \sin x$.

Solution:

Using the product rule and the formulas above, we obtain

$f'(x) &= x^2 (-\!\sin x) + 2x \cos x + \cos x\\&= - x^2 \sin x + 2x \cos x + \cos x.$

Example 3:

Find $dy/dx$ if $y = \frac {\cos x}{1-\tan x}$ . What is the slope of the tangent line at $x = \pi/3$?

Solution:

Using the quotient rule and the formulas above, we obtain

$\frac {dy}{dx} &= \frac {(1-\tan x)(-\!\sin x) - (\cos x)(-\!\sec^2 x)} {(1-\tan x)^2}\\ &= \frac {-\!\sin x + \tan x \sin x + \cos x \sec^2 x}{(1- \tan x)^2}\\ &= \frac {-\!\sin x + \tan x \sin x + \sec x}{(1- \tan x)^2}$

To calculate the slope of the tangent line, we simply substitute $x = \pi/3$:

$\frac {dy}{dx} \Bigg|_{x = \frac {\pi}{3}} = \frac {-\!\sin( \frac{\pi}{3} ) + \tan ( \frac{\pi}{3} ) \sin ( \frac{\pi}{3} ) + \sec( \frac{\pi}{3} )} {(1 - \tan( \frac{\pi}{3} ))^2}.$

We finally get the slope to be approximately

$\frac {dy}{dx} \Bigg|_{x = \frac {\pi}{3}} = 4.9.$

Example 4:

If $y = \sec x$, find $y'' (\pi/3)$.

Solution:

$y' &= \sec x \tan x\\y'' &= \sec x (\sec^2 x) + (\sec x \tan x) \tan x\\&= \sec^3 x + \sec x \tan^2 x.$

Substituting for $x = \pi/3$,

$y'' &= \sec^3 \left(\frac{\pi}{3}\right) + \sec \left(\frac{\pi}{3}\right) \tan^2 \left(\frac{\pi}{3}\right)\\&= (2)^3 + (2)(\sqrt{3})^2\\ &= 8 + (2)(3)\\&= 14.$

Thus $y'' (\pi/3) = 14$.

For examples of finding the derivatives of trigonometric functions (4.4), see Math Video Tutorials by James Sousa, The Derivative of Sine and Cosine (9:21).

## Review Questions

Find the derivative $y'$ of the following functions:

1. $y = x \sin x + 2$
2. $y = x^2 \cos x - x \tan x - 1$
3. $y = \sin^2 x$
4. $y = \frac {\sin x-1}{\sin x+1}$
5. $y = \frac {\cos x+\sin x} {\cos x - \sin x}$
6. $y = \frac {\sqrt {x}}{\tan x} + 2$
7. $y = \csc x \sin x + x$
8. $y = \frac {\sec x} {\csc x}$
9. If $y = \csc x$, find $y'' (\pi/6).$
10. Use the definition of the derivative to prove that $\frac {d}{dx}[\cos x] = - \sin x.$

1. $y' = x \cos x + \sin x$
2. $y' = 2x \cos x - x^2 \sin x - \tan x - x \sec^2 x$
3. $y' = 2 \cos x \sin x$
4. $y' =\frac {2 \cos x} {(\sin x+1)^2}$
5. $y' = 1 +\left ( \frac{\tan x+1}{1-\tan x} \right )^2$
6. $y' =\frac {\cot} {2 \sqrt {x}} - \sqrt {x} \csc^2 x$
7. $y' = 1$
8. $y' = \sec^2 x$
9. $y(\pi/6) = 14$

$\frac{d \cos x}{dx} &= \lim_{h \to 0} \frac{\cos(x +h) - \cos x}{h}\\&= \lim_{h \to 0} \frac{(\cos x \cos h - \sin x \sin h) - \cos x}{h} \ \text{(From the identity for cosine angle sum).}\\&= \lim_{h \to 0} \left [ \cos x \left ( \frac{\cos h - 1}{h} \right ) - \sin x \left ( \frac{\sin h}{h} \right ) \right]\\&= \lim_{h \to 0} \cos x \left ( \frac{\cos h - 1}{h} \right ) - \lim_{h \to 0} \sin x \left( \frac{\sin h}{h} \right )\\&= \cos x \lim_{h \to 0} \left ( \frac{\cos h - 1}{h} \right ) - \sin x \lim_{h \to 0} \left ( \frac{\sin h}{h} \right )\\&= \cos x (0) - \sin x(1) \ \text{(with a note that these two limits are demonstrated in the multimedia links).}\\&= - \sin x$

Feb 23, 2012

Feb 26, 2015