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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 \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*}.

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*}

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