Differentiate the following w.r.t. as indicated :
`secx" w.r.t. "cosecx`

Answer

To differentiate \( \sec x \) with respect to \( \csc x \), we will use the chain rule and implicit differentiation. Let's denote:

- \( u = \sec x \)
- \( v = \csc x \)

We want to find \( \frac{du}{dv} \).

### Step 1: Differentiate \( u \) and \( v \) with respect to \( x \)

1. **Differentiate \( u = \sec x \)**:
\[ \frac{du}{dx} = \sec x \tan x \]

2. **Differentiate \( v = \csc x \)**:
\[ \frac{dv}{dx} = -\csc x \cot x \]

### Step 2: Use the chain rule to find \( \frac{du}{dv} \)

Using the chain rule, we have:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

Substituting the derivatives we found:
\[ \frac{du}{dv} = \frac{\sec x \tan x}{-\csc x \cot x} \]

### Step 3: Simplify the expression

Now, we simplify \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{\sec x \tan x}{-\csc x \cot x} = -\frac{\sec x \tan x}{\csc x \cot x} \]

Recall that:
- \( \sec x = \frac{1}{\cos x} \)
- \( \tan x = \frac{\sin x}{\cos x} \)
- \( \csc x = \frac{1}{\sin x} \)
- \( \cot x = \frac{\cos x}{\sin x} \)

Substituting these identities:
\[ \frac{du}{dv} = -\frac{\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}}{\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}} = -\frac{\frac{\sin x}{\cos^2 x}}{\frac{\cos x}{\sin^2 x}} = -\frac{\sin^3 x}{\cos^3 x} \]

Thus, the final answer is:
\[ \frac{du}{dv} = -\frac{\sin^3 x}{\cos^3 x} \]