Question

1-22 Differentiate.

20.\(g\left( z \right) = \frac{z}{{\sec z + \tan z}}\)

Short answer

The differentiation of the function \(g\left( z \right) = \frac{z}{{\sec z + \tan z}}\) is \(g'\left( z \right) = \frac{{1 - z\sec z}}{{\sec z + \tan z}}\).

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 \(g\left( z \right) = \frac{z}{{\sec z + \tan z}}\). Differentiate the function w.r.t \(z\) by using the derivatives of trigonometric functions and the quotient rule.

\(\begin{aligned}\frac{{d\left( {g\left( z \right)} \right)}}{{dz}} &= \frac{d}{{dz}}\left( {\frac{z}{{\sec z + \tan z}}} \right)\\ &= \frac{{\left( {\sec z + \tan z} \right)\frac{d}{{dz}}\left( z \right) - z\frac{d}{{dz}}\left( {\sec z + \tan z} \right)}}{{{{\left( {\sec z + \tan z} \right)}^2}}}\\ &= \frac{{\left( {\sec z + \tan z} \right)\left( 1 \right) - z\left( {\sec z\tan z + {{\sec }^2}z} \right)}}{{{{\left( {\sec z + \tan z} \right)}^2}}}\\ &= \frac{{\sec z + \tan z - z\sec z\left( {\sec z + \tan z} \right)}}{{{{\left( {\sec z + \tan z} \right)}^2}}}\\ &= \frac{{\left( {\sec z + \tan z} \right)\left( {1 - z\sec z} \right)}}{{{{\left( {\sec z + \tan z} \right)}^2}}}\\ &= \frac{{1 - z\sec z}}{{\sec z + \tan z}}\end{aligned}\)

Thus, the derivative of the function \(g\left( z \right) = \frac{z}{{\sec z + \tan z}}\) is \(g'\left( z \right) = \frac{{1 - z\sec z}}{{\sec z + \tan z}}\).