Question

1-22 Differentiate.

7. \(y = sec\theta tan\theta \)

Short answer

The differentiation of the function \(y = \sec \theta \tan \theta \) is \(y' = \sec \theta \left( {2{{\sec }^2}\theta - 1} \right)\).

Step-by-step solution

 Step 1: Write the formula of the derivatives of trigonometric functions and the product rule

\(\begin{aligned}\frac{d}{{dx}}\left( {\tan x} \right) &= {\sec ^2}x\\\frac{d}{{dx}}\left( {\sec x} \right) &= \sec x\tan x\end{aligned}\)

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

 Step 2: Find the differentiation of the function

Consider the function \(y = \sec \theta \tan \theta \). Differentiate the function w.r.t \(\theta \) by using the derivatives of trigonometric functions and the product rule.

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

Furthermore, use\(1 + {\tan ^2}\theta = {\sec ^2}\theta \).

\(\begin{aligned}y' &= \sec \theta \left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right)\\ &= \sec \theta \left( {2{{\sec }^2}\theta - 1} \right)\end{aligned}\)

Thus, the derivative of the function \(y = \sec \theta \tan \theta \) is \(y' = \sec \theta \left( {2{{\sec }^2}\theta - 1} \right)\).