\( [ ( \tan x + \cos x ) ( \sin x + \cot x ) ] ^ { \prime } \)

To differentiate the expression \( y = [ ( \tan x + \cos x ) ( \sin x + \cot x ) ] \), we will use the product rule. The product rule states that if \( y = u v \), then \( y' = u' v + u v' \). Let: \[ u = \tan x + \cos x \] \[ v = \sin x + \cot x \] We need to find \( u' \) and \( v' \). ### Step 1: Differentiate \( u \) \[ u' = \frac{d}{dx}(\tan x) + \frac{d}{dx}(\cos x) \] \[ u' = \sec^2 x - \sin x \] ### Step 2: Differentiate \( v \) \[ v' = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cot x) \] \[ v' = \cos x - \csc^2 x \] ### Step 3: Apply the product rule Now we can apply the product rule: \[ y' = u' v + u v' \] Substituting \( u \), \( u' \), \( v \), and \( v' \): \[ y' = (\sec^2 x - \sin x)(\sin x + \cot x) + (\tan x + \cos x)(\cos x - \csc^2 x) \] ### Step 4: Simplify the expression Now we can expand both terms: 1. For the first term: \[ (\sec^2 x - \sin x)(\sin x + \cot x) = \sec^2 x \sin x + \sec^2 x \cot x - \sin^2 x - \sin x \cot x \] 2. For the second term: \[ (\tan x + \cos x)(\cos x - \csc^2 x) = \tan x \cos x - \tan x \csc^2 x + \cos^2 x - \cos x \csc^2 x \] ### Final expression Combining these results gives us: \[ y' = \sec^2 x \sin x + \sec^2 x \cot x - \sin^2 x - \sin x \cot x + \tan x \cos x - \tan x \csc^2 x + \cos^2 x - \cos x \csc^2 x \] This is the derivative of the given expression. Thus, the final answer is: \[ y' = \sec^2 x \sin x + \sec^2 x \cot x - \sin^2 x - \sin x \cot x + \tan x \cos x - \tan x \csc^2 x + \cos^2 x - \cos x \csc^2 x \]