Question

Find the derivative of the function:

32. \(J\left( \theta \right) = {\tan ^2}\left( {n\theta } \right)\)

Short answer

The derivative of \(J\left( \theta \right)\) is \(2n\tan \left( {n\theta } \right){\sec ^2}\left( {n\theta } \right)\).

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the given function

Differentiate \(J\left( \theta \right)\) with respect to \(\theta \) as follows:

\(\begin{aligned}J\left( \theta \right) &= \frac{d}{{d\theta }}\left( {{{\tan }^2}\left( {n\theta } \right)} \right)\\ &= \frac{d}{{d\theta }}{\left( {\tan \left( {n\theta } \right)} \right)^2}\\ &= 2\tan \left( {n\theta } \right)\frac{d}{{d\theta }}\left( {\tan \left( {n\theta } \right)} \right)\\ &= 2\tan \left( {n\theta } \right)\left( {{{\sec }^2}\left( {n\theta } \right)} \right)\frac{d}{{d\theta }}\left( {n\theta } \right)\\ &= 2\tan \left( {n\theta } \right)\left( {{{\sec }^2}\left( {n\theta } \right)} \right)\left( n \right)\\ &= 2n\tan \left( {n\theta } \right){\sec ^2}\left( {n\theta } \right)\end{aligned}\)

Hence, the derivative of \(J\left( \theta \right)\) is \(2n\tan \left( {n\theta } \right){\sec ^2}\left( {n\theta } \right)\).