Question

1-22 Differentiate.

15.\(y = \frac{x}{{2 - \tan x}}\)

Short answer

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

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 \(y = \frac{x}{{2 - \tan x}}\). Differentiate the function w.r.t \(x\) by using the derivatives of trigonometric functions and the quotient rule.

\(\begin{aligned}\frac{{d\left( y \right)}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{x}{{2 - \tan x}}} \right)\\ &= \frac{{\left( {2 - \tan x} \right)\frac{d}{{dx}}\left( x \right) - x\frac{d}{{dx}}\left( {2 - \tan x} \right)}}{{{{\left( {2 - \tan x} \right)}^2}}}\\ &= \frac{{\left( {2 - \tan x} \right)\left( 1 \right) - x\left( { - {{\sec }^2}x} \right)}}{{{{\left( {2 - \tan x} \right)}^2}}}\\ &= \frac{{2 - \tan x + x{{\sec }^2}x}}{{{{\left( {2 - \tan x} \right)}^2}}}\end{aligned}\)

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