Question

Find the derivative of the function:

25. \(y = {e^{\tan \theta }}\)

Short answer

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

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 \(y\) with respect to \(\theta \) as follows:

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

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