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2.4: Derivatives of Trigonometric Functions

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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.

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 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.

Review Answers

  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

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Feb 23, 2012

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Sep 01, 2014
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