Derivatives of trigonometric functions

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We use the product and quotient 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 $\sin (x)$ and $\cos (x)$ . We know that $\ddx \sin (x) = \cos (x).$ and $\ddx \cos (x) = -\sin (x).$ Armed with this fact we will discover the derivatives of all of the standard trigonometric functions.

Next we have:

The derivative of tangent $\ddx \tan (x) = \sec ^2(x).$

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

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

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

Finally, we have:

The derivative of secant $\ddx \sec (x) = \sec (x)\tan (x).$

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

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

The Derivatives of Trigonometric Functions

$\ddx \sin (x) = \cos (x)$ .
$\ddx \cos (x) = -\sin (x)$ .
$\ddx \tan (x) = \sec ^2(x)$ .
$\ddx \sec (x) = \sec (x)\tan (x)$ .
$\ddx \csc (x) = -\csc (x)\cot (x)$ .
$\ddx \cot (x) = -\csc ^2(x)$ .

Compute: $\eval {\ddx ( \csc (x) \cot (x) )}_{x=\frac {\pi }{3}}$

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