Differentiate the following w.r.t. x :
`(cosx)/(tanx),xgt0`

Answer

To differentiate the function \( y = \frac{\cos x}{\tan x} \) with respect to \( x \), we will use the quotient rule. The quotient rule states that if you have a function \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = \cos x \) and \( v = \tan x \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = \cos x \)
- \( v = \tan x \)

### Step 2: Differentiate \( u \) and \( v \)
Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = -\sin x \)
- \( \frac{dv}{dx} = \sec^2 x \) (since \( \tan x = \frac{\sin x}{\cos x} \) and its derivative is \( \sec^2 x \))

### Step 3: Apply the Quotient Rule
Now we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{\tan x (-\sin x) - \cos x (\sec^2 x)}{\tan^2 x} \]

### Step 4: Simplify the Expression
Now we simplify the expression:

1. The first term in the numerator is \( -\tan x \sin x \).
2. Recall that \( \tan x = \frac{\sin x}{\cos x} \), so \( -\tan x \sin x = -\frac{\sin^2 x}{\cos x} \).
3. The second term is \( -\cos x \sec^2 x = -\cos x \cdot \frac{1}{\cos^2 x} = -\frac{1}{\cos x} \).

Thus, we have:

\[ \frac{dy}{dx} = \frac{-\frac{\sin^2 x}{\cos x} - \frac{1}{\cos x}}{\tan^2 x} \]

Combine the terms in the numerator:

\[ \frac{dy}{dx} = \frac{-\frac{\sin^2 x + 1}{\cos x}}{\tan^2 x} \]

### Step 5: Rewrite in Terms of \( \tan x \)
Since \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \), we can rewrite the expression:

\[ \frac{dy}{dx} = -\frac{\sin^2 x + 1}{\cos x} \cdot \frac{\cos^2 x}{\sin^2 x} \]

This can be simplified further, but it is already in a usable form.

### Final Answer
Thus, the derivative of \( y = \frac{\cos x}{\tan x} \) with respect to \( x \) is:

\[ \frac{dy}{dx} = -\frac{\sin^2 x + 1}{\sin^2 x \cos x} \]

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