<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

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 \begin{align*}h\end{align*} is measured in radians,

\begin{align*} \lim_{h \to 0} \frac{\sin h} {h} = 1\end{align*} and \begin{align*}\lim_{h \to 0} \frac{1 - \cos h} {h} = 0.\end{align*}

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

\begin{align*} \frac{d} {dx} [\sin x] = \lim_{h \to 0} \frac{\sin(x + h) - \sin x} {h}\end{align*}

Since

\begin{align*}\sin(\alpha + \beta) = \sin \alpha\ \cos \beta + \cos \alpha\ \sin \beta.\end{align*}

The derivative becomes

\begin{align*}\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.\end{align*}

Therefore,

\begin{align*}\frac {d}{dx} [\sin x] = \cos x.\end{align*}

It will be left as an exercise to prove that

\begin{align*}\frac {d}{dx}[\cos x] = -\!\sin x.\end{align*}

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

Derivatives of Trigonometric Functions

\begin{align*}\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}\end{align*}

Keep in mind that for all the derivative formulas for the trigonometric functions, the argument \begin{align*}x\end{align*} is measured in radians.

Example 1:

Show that \begin{align*} \frac {d}{dx} [\tan x] = \sec^2 x.\end{align*}

Solution:

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

Since

\begin{align*} \tan x =\frac{\sin x}{\cos x},\end{align*}

then

\begin{align*}\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}\end{align*}

Example 2:

Find \begin{align*}f'(x) \mathrm{~if~} f(x) = x^{2} \cos x + \sin x\end{align*}.

Solution:

Using the product rule and the formulas above, we obtain

\begin{align*}f'(x) &= x^2 (-\!\sin x) + 2x \cos x + \cos x\\ &= - x^2 \sin x + 2x \cos x + \cos x.\end{align*}

Example 3:

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

Solution:

Using the quotient rule and the formulas above, we obtain

\begin{align*}\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}\end{align*}

To calculate the slope of the tangent line, we simply substitute \begin{align*}x = \pi/3\end{align*}:

\begin{align*} \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}.\end{align*}

We finally get the slope to be approximately

\begin{align*} \frac {dy}{dx} \Bigg|_{x = \frac {\pi}{3}} = 4.9.\end{align*}

Example 4:

If \begin{align*}y = \sec x\end{align*}, find \begin{align*}y'' (\pi/3)\end{align*}.

Solution:

\begin{align*}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.\end{align*}

Substituting for \begin{align*}x = \pi/3\end{align*},

\begin{align*}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.\end{align*}

Thus \begin{align*}y'' (\pi/3) = 14\end{align*}.

Multimedia Links

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 \begin{align*}y'\end{align*} of the following functions:

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

Image Attributions

Show Hide Details
Description
Subjects:
Grades:
Date Created:
Feb 23, 2012
Last Modified:
Oct 30, 2015
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the section. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.1.Calculus.2.4

Original text