Question

differentiate.
$y = \frac{3 - \sec x}{\tan x}$
$y = \square$

Explanation:

Step1: Apply quotient rule

Let \( u = 3 - \sec x \), \( v = \tan x \). Quotient rule: \( y' = \frac{u'v - uv'}{v^2} \)

Step2: Compute derivatives of u and v

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

Step3: Substitute into quotient rule

\( y' = \frac{(-\sec x \tan x)(\tan x) - (3 - \sec x)(\sec^2 x)}{\tan^2 x} \)

Step4: Simplify numerator

\( = \frac{-\sec x \tan^2 x - 3\sec^2 x + \sec^3 x}{\tan^2 x} \)

Step5: Use \( \tan^2 x = \sec^2 x - 1 \)

\( = \frac{-\sec x (\sec^2 x - 1) - 3\sec^2 x + \sec^3 x}{\tan^2 x} = \frac{-\sec^3 x + \sec x - 3\sec^2 x + \sec^3 x}{\tan^2 x} = \frac{\sec x - 3\sec^2 x}{\tan^2 x} \)

Step6: Factor numerator and simplify

\( = \frac{\sec x (1 - 3\sec x)}{\tan^2 x} = \frac{\frac{1}{\cos x}(1 - \frac{3}{\cos x})}{\frac{\sin^2 x}{\cos^2 x}} = \frac{(1 - 3\sec x)\cos x}{\sin^2 x} = \frac{\cos x - 3}{\sin^2 x} \)

Answer:

\( \frac{\cos x - 3}{\sin^2 x} \)