Question

1-22: Differentiate.

4. \(y = 2\sec x - \csc x\)

Short answer

The required value is \(y' = 2\sec x\tan x + \csc x\cot x\).

Step-by-step solution

Write the formula of the derivatives of trigonometric functions

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

Find the differentiation of the function

Consider the function \(y = 2\sec x - \csc x\). Differentiate the function w.r.t \(x\) by using the derivatives of trigonometric functions.

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

Thus, the derivative of the function \(y = 2\sec x - \csc x\) is \(y' = 2\sec x\tan x + \csc x\cot x\).