Question

1. - / 5.55 points differentiate. $y = \frac{2x}{1 - \tan x}$ $y = \square$

Explanation:

Step1: Identify numerator and denominator

Let \( u = 2x \), \( v = 1 - \tan x \)

Step2: Compute derivatives of u and v

\( u' = 2 \), \( v' = -\sec^2 x \)

Step3: Apply quotient rule \( \frac{u'v - uv'}{v^2} \)

\( y' = \frac{2(1 - \tan x) - 2x(-\sec^2 x)}{(1 - \tan x)^2} \)

Step4: Simplify numerator

\( y' = \frac{2 - 2\tan x + 2x\sec^2 x}{(1 - \tan x)^2} \)

Answer:

\( \frac{2 - 2\tan x + 2x\sec^2 x}{(1 - \tan x)^2} \)