Derivatives of trigonometric functions

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We use the chain rule to unleash the derivatives of the trigonometric functions.

Up until this point of the course we have been ignoring a large class of functions: Trigonometric functions other than $\sin (x)$ . We know that $\frac {d}{dx} \sin (x) = \cos (x).$ Armed with this fact we will discover the derivatives of all of the standard trigonometric functions.

The derivative of cosine $\frac {d}{dx} \cos (x) = -\sin (x).$

Recall that

$\cos (x) = \sin \left (\frac {\pi }{2}-x\right )$ , and
$\sin (x) = \cos \left (\frac {\pi }{2}-x\right )$ .

Now

$\begin{align*} \frac{d}{dx} \cos(x) &= \frac{d}{dx} \sin\left(\frac{\pi}{2}-x\right)\\ &=-\cos\left(\frac{\pi}{2}-x\right) \\ &= -\sin(x). \end{align*}$

Compute: $\bigg [ \frac {d}{dx} \cos \left ( \frac {x^3}{2} \right ) \bigg ]_{x=\sqrt [3]{\pi }}$

Now that we know the derivative of cosine, we may combine this with the chain rule, so we have that $\frac {d}{dx} \cos \left ( \frac {x^3}{2} \right ) = \answer [given]{\frac {3 x^2}{2}} \left (- \sin \left ( \frac {x^3}{2} \right ) \right )$ and so $\bigg [ \frac {d}{dx} \cos \left ( \frac {x^3}{2} \right ) \bigg ]_{x=\sqrt [3]{\pi }}$ $\begin{align*} &= \bigg[ \left( \frac{3}{2} x^2 \left(- \sin \left( \frac{x^3}{2} \right) \right) \right) \bigg]_{x=\sqrt[3]{\pi}} \\ &= - \frac{3}{2}(\sqrt[3]{\pi})^2 \sin \left( \frac{\pi}{2} \right) \\ &= -\frac{3}{2} \pi^{\frac{2}{3}} \cdot \answer[given]{1} \\ &=\answer[given]{\frac{-3 \pi^{\frac{2}{3}}}{2}}. \end{align*}$

Next we have:

The derivative of tangent $\frac {d}{dx} \tan (x) = \sec ^2(x).$

We’ll rewrite $\tan (x)$ as $\frac {\sin (x)}{\cos (x)}$ and use the quotient rule. Write with me: $\begin{align*} \frac{d}{dx}\tan(x) &= \frac{d}{dx}\frac{\sin(x)}{\cos(x)}\\ &=\frac{\cos^2(x) + \answer[given]{\sin^2(x)}}{\cos^2(x)}\\ &=\frac{\answer[given]{1}}{\cos^2(x)}\\ &=\sec^2(x). \end{align*}$

Compute: $\frac {d}{dx} \left ( \frac {5x \tan (x)}{x^2 - 3} \right )$

Applying the quotient rule, and the product rule, and the derivative of tangent: $\begin{align*} \frac{d}{dx} &\left( \frac{5x \tan(x)}{x^2 - 3} \right) \\ &= \frac{(x^2 - 3) \cdot \frac{d}{dx}(\answer[given]{5x \tan(x)}) - 5x \tan(x) \cdot \frac{d}{dx} (\answer[given]{x^2 - 3})}{(x^2 - 3)^2} \\ &= \frac{(x^2 - 3)(5 \tan(x) + 5x \answer[given]{\sec^2(x)}) - 5x \tan(x) \cdot 2x}{(x^2 - 3)^2} \\ &= \frac{5(x^2-3)(\tan(x)+x \sec^2(x)) - 10x^2 \tan(x)}{(x^2-3)^2} \end{align*}$

Finally, we have:

The derivative of secant $\frac {d}{dx} \sec (x) = \sec (x)\tan (x).$

We’ll rewrite $\sec (x)$ as $(\cos (x))^{-1}$ and use the power rule and the chain rule. Write $\begin{align*} \frac{d}{dx} \sec(x) &= \frac{d}{dx}(\cos (x))^{-1}\\ &=-1(\cos(x))^{-2}(\answer[given]{-\sin(x)}) \\ &= \frac{\sin(x)}{\cos^2(x)} \\ &= \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} \\ &= \sec(x)\tan(x). \end{align*}$

The derivatives of the cotangent and cosecant are similar and left as exercises. Putting this all together, we have:

The Derivatives of Trigonometric Functions

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

Compute: $\bigg [ \frac {d}{dx} ( \csc (x) \cot (x) ) \bigg ]_{x=\frac {\pi }{3}}$

Applying the product rule and the facts above, we know that $\frac {d}{dx} ( \csc (x) \cot (x) ) = - \csc ^3(x) - \cot ^2(x)\answer [given]{\csc (x)}$ and so $\bigg [ \frac {d}{dx} ( \csc (x) \cot (x) ) \bigg ]_{x=\frac {\pi }{3}}$ $\begin{align*} &= \bigg[ - \csc^3(x) - \cot^2(x) \answer[given]{\csc(x)} \bigg]_{x=\frac{\pi}{3}} \\ &= - \frac{8}{3 \sqrt{3}} - \frac{1}{3}\cdot \answer[given]{2/\sqrt{3}} \end{align*}$

When working with derivatives of trigonometric functions, we suggest you use radians for angle measure. For example, while $\sin \left ((90^\circ \right )^2) = \sin \left (\left (\frac {\pi }{2}\right )^2\right ),$ one must be careful with derivatives as $\bigg [ \frac {d}{dx} \sin \left (x^2\right ) \bigg ]_{x=90^\circ } \ne \underbrace {2\cdot 90\cdot \cos (90^2)}_{\text {incorrect}}$ Alternatively, one could think of $x^\circ$ as meaning $\frac {x\cdot \pi }{180}$ , as then $90^\circ = \frac {90\cdot \pi }{180} = \frac {\pi }{2}$ . In this case $2\cdot 90^\circ \cdot \cos ((90^\circ )^2) = 2\cdot \frac {\pi }{2}\cdot \cos \left (\left (\frac {\pi }{2}\right )^2\right ).$