Question

7-52: Find the derivative of the function.

13. \(f\left( \theta \right) = \cos \left( {{\theta ^2}} \right)\)

Short answer

The derivative of the function is \(f'\left( \theta \right) = - 2\theta \sin \left( {{\theta ^2}} \right)\).

Step-by-step solution

The Chain Rule

The chain rule is defined as:

\(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\)

The Leibniz notationis defined as:

\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

Find the derivative of the function

The inner function is \(u = g\left( \theta \right) = {\theta ^2}\) , and the outer function is \(y = f\left( u \right) = \cos u\).

Use the chain rule to obtain the derivative of the function as shown below

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

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