Question

1-22: Differentiate.

5. \(h\left( \theta \right) = {\theta ^2}\sin \theta \)

Short answer

The required value is \(h'\left( \theta \right) = \theta \left( {\theta \cos \theta + 2\sin \theta } \right)\).

Step-by-step solution

Write the formula of the derivatives of trigonometric functions, the product rule and the power function 

\(\frac{d}{{dx}}\left( {\sin x} \right) = \cos x\)

The product rule: If \(f\) and \(g\) are differentiable functions, then\(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\).

The Power Rule:\(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)where \(n\) is a positive integer

Find the differentiation of the function

Consider the function \(h\left( \theta \right) = {\theta ^2}\sin \theta \). Differentiate the function w.r.t \(\theta \) by usingthe derivatives of trigonometric functions, the product rule and the power function.

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

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