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
\[ \ddx \sin (x) = \cos (x). \]
Armed with this fact we will discover the derivatives of all of the standard trigonometric functions.
Compute:
\[ \eval {\ddx \cos \left ( \frac {x^3}{2} \right )}_{x=\sqrt [3]{\pi }} \]
Now that we know the derivative of cosine, we may combine this with the
chain rule, so we have that
\[ \ddx \cos \left ( \frac {x^3}{2} \right ) = \answer [given]{\frac {3 x^2}{2}} \left (- \sin \left ( \frac {x^3}{2} \right ) \right ) \]
and so
\begin{align*} \eval {\ddx \cos \left ( \frac {x^3}{2} \right )}_{x=\sqrt [3]{\pi }} &= \eval {\left ( \frac {3}{2} x^2 \left (- \sin \left ( \frac {x^3}{2} \right ) \right ) \right )}_{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
\[ \ddx \tan (x) = \sec ^2(x). \]
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
tangent:
\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 and 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*}
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 \[ \eval {\ddx \sin \left (x^2\right )}_{x=90^\circ } \ne \underbrace {2\cdot 90\cdot \cos (90^2)}_{\text {incorrect}} \]
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 \[ 2\cdot 90^\circ \cdot \cos ((90^\circ )^2) = 2\cdot \frac {\pi }{2}\cdot \cos \left (\left (\frac {\pi }{2}\right )^2\right ). \]