Question

1-22: Differentiate.

21.\(f\left( \theta \right) = \theta \cos \theta \sin \theta \)

Short answer

The derivative of the function is \(f'\left( \theta \right) = \frac{1}{2}\sin \theta + \theta \cos 2\theta \).

Step-by-step solution

 Derivative of Trigonometric Functions

\(\begin{aligned}\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\csc x} \right) &= - \csc x\cot x\\\frac{d}{{dx}}\left( {\cos x} \right) &= - \sin x\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\sec x} \right) &= \sec x\tan x\\\frac{d}{{dx}}\left( {\tan x} \right) &= {\sec ^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{{dx}}\left( {\cot x} \right) &= - {\csc ^2}x\end{aligned}\)

Differentiate the function

Product Rule can be extended to the product of three functions is given by \({\left( {fgh} \right)^\prime } = f'gh + fg'h + fgh'\).

Use the above rule to differentiate the function as shown below:

\(\begin{aligned}f'\left( \theta \right) &= \frac{d}{{d\theta }}\left( {\theta \cos \theta \sin \theta } \right)\\ &= \left( {\cos \theta \sin \theta } \right)\frac{d}{{d\theta }}\left( \theta \right) + \left( {\theta \sin \theta } \right)\frac{d}{{d\theta }}\left( {\cos \theta } \right) + \left( {\theta \cos \theta } \right)\frac{d}{{d\theta }}\left( {\sin \theta } \right)\\ &= \left( {\sin \theta \cos \theta } \right)\left( 1 \right) + \left( {\theta \sin \theta } \right)\left( { - \sin \theta } \right) + \left( {\theta \cos \theta } \right)\left( {\cos \theta } \right)\\ &= \sin \theta \cos \theta - \theta {\sin ^2}\theta + \theta {\cos ^2}\theta \\ &= \sin \theta \cos \theta + \theta \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)\\ &= \frac{1}{2}\sin \theta + \theta \cos 2\theta \end{aligned}\)

Thus, the derivative of the function is \(f'\left( \theta \right) = \frac{1}{2}\sin \theta + \theta \cos 2\theta \).