`d/(dx)(cotx)` is :

Options

A. `tanx`
B. `-cosec^(2)x`
C. `-cosecxcotx`
D. None of these

Answer

To find the derivative of \( \cot x \), we can use the following steps:

### Step 1: Recall the definition of \( \cot x \)
The cotangent function can be expressed as:
\[ \cot x = \frac{\cos x}{\sin x} \]

### Step 2: Use the quotient rule
To differentiate \( \cot x \), we apply the quotient rule, which states that if \( y = \frac{u}{v} \), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, let \( u = \cos x \) and \( v = \sin x \).

### Step 3: Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)
We know:
\[ \frac{du}{dx} = -\sin x \quad \text{(derivative of } \cos x\text{)} \]
\[ \frac{dv}{dx} = \cos x \quad \text{(derivative of } \sin x\text{)} \]

### Step 4: Substitute into the quotient rule
Now substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule:
\[ \frac{d}{dx}(\cot x) = \frac{\sin x (-\sin x) - \cos x \cos x}{\sin^2 x} \]

### Step 5: Simplify the expression
This simplifies to:
\[ \frac{d}{dx}(\cot x) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \frac{d}{dx}(\cot x) = \frac{-1}{\sin^2 x} \]

### Step 6: Final result
Thus, we have:
\[ \frac{d}{dx}(\cot x) = -\csc^2 x \]

### Summary
The derivative of \( \cot x \) is:
\[ \frac{d}{dx}(\cot x) = -\csc^2 x \]