Question

1-22 Differentiate.

17.\(f\left( w \right) = \frac{{1 + \sec w}}{{1 - \sec w}}\)

Short answer

The differentiation of the function \(f\left( w \right) = \frac{{1 + \sec w}}{{1 - \sec w}}\) is \(f'\left( w \right) = \frac{{2\sec w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}}\).

Step-by-step solution

Write the formula of the derivatives of trigonometric functions and the quotient 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 Quotient rule: If \(f\) and \(g\) are differentiable functions, then\(\frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right) + f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\).

Find the differentiation of the function

Consider the function \(f\left( w \right) = \frac{{1 + \sec w}}{{1 - \sec w}}\). Differentiate the function w.r.t \(w\) by using the derivatives of trigonometric functions and the quotient rule.

\(\begin{aligned}\frac{{d\left( {f\left( w \right)} \right)}}{{dw}} &= \frac{d}{{dw}}\left( {\frac{{1 + \sec w}}{{1 - \sec w}}} \right)\\ &= \frac{{\left( {1 - \sec w} \right)\frac{d}{{dw}}\left( {1 + \sec w} \right) - \left( {1 + \sec w} \right)\frac{d}{{dw}}\left( {1 - \sec w} \right)}}{{{{\left( {1 - \sec w} \right)}^2}}}\\ &= \frac{{\left( {1 - \sec w} \right)\left( {\sec w\tan w} \right) - \left( {1 + \sec w} \right)\left( { - \sec w\tan w} \right)}}{{{{\left( {1 - \sec w} \right)}^2}}}\\ &= \frac{{\sec w\tan w - {{\sec }^2}w\tan w + \sec w\tan w + {{\sec }^2}w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}}\\ &= \frac{{2\sec w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}}\end{aligned}\)

Thus, the derivative of the function \(f\left( w \right) = \frac{{1 + \sec w}}{{1 - \sec w}}\) is \(f'\left( w \right) = \frac{{2\sec w\tan w}}{{{{\left( {1 - \sec w} \right)}^2}}}\).