Derivatives of Trigonometric Functions

Videos

Worksheets

Derivatives of Trigonometric Functions

As we’ll learn in this section, there are several ways to compute derivatives of trigonometric functions. First we’ll calculate the slope of the tangent line to the graph at several points, and use our answers to plot the graph of the derivative. Second, we’ll show how Quotient Rule can be used to compute the derivatives of trigonometric identities. But, the takeaway from this section is that you should really just memorize the derivatives of at least the six basic trigonometric identities. In the end, doing so will save you a lot of time.

Consider the graph of the function $y = \sin(x)$ .

graph of sin(x)

Recall that the derivative of a function at a certain point gives the slope of the line tangent to the function at that point. Working backwards, we’ll look at the slopes of the tangent lines to $y = \sin(x)$ , and then try to determine the derivative function.

graph of sin(x) with tangent lines drawn

Drawing the tangent lines and recording the slopes at $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ , and so on, we can see that the derivative of this function should be 1 at 0, 0 at $\frac{\pi}{2}$ , -1 at $\pi$ , 0 at $\frac{3\pi}{2}$ , 1 at $2\pi$ etc.

the points (0,1), ( $\frac{\pi}{2}$ ,0), ( $\pi$ ,-1), ( $\frac{3\pi}{2}$ ,0) etc. of the derivative function

Note that this has a striking resemblance to the graph of $y = \cos(x)$ .

graph of cos(x)

So we have that $\frac{d}{dx} \sin(x) = \cos(x)$ .

In the same way, it can be shown that $\frac{d}{dx} \cos(x) = -\sin(x)$ .

When considering the derivatives of the other trigonometric functions, the quotient rule is useful. If you do not know the quotient rule, you can ignore the justification of $\frac{d}{dx} \tan(x) = \sec^2(x)$ and come back to look at this section after you have been introduced to the quotient rule.

Recall that $\tan(x) = \frac{\sin(x)}{\cos(x)}$ . So we have that:

$\frac{d}{dx}\tan(x)=\frac{d}{dx}\frac{\sin(x)}{\cos(x)}=\frac{\cos(x)\cdot \cos(x)-\sin(x)\cdot (-\sin(x))}{\cos^2(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}=\sec^2(x)$

(Using the trigonometric identity $\cos^2(x)+\sin^2(x)=1$ , and the definition $\frac{1}{\cos(x)}=\sec(x)$ )

The derivatives for $\cot(x), \csc(x),$ and $\sec(x)$ can also be derived using the quotient rule.

$\frac{d}{dx} \sin(x) = \cos(x)$
$\frac{d}{dx} \cos(x) = -\sin(x)$
$\frac{d}{dx} \tan(x) = \sec^2(x)$
$\frac{d}{dx} \cot(x) = -\csc^2(x)$
$\frac{d}{dx} \sec(x) = \sec(x)\tan(x)$
$\frac{d}{dx} \csc(x) = -\csc(x)\cot(x)$

A trick to remembering where the negative signs are: whenever the “sine” function is present in the derivative, include a “negative sign” in front of it. (You also include “-” before cosecant; because $\csc(x) = 1/\sin(x)$ .)