Differentiate the following w.r.t. as indicated :
`tanx" w.r.t. "cosx`

Answer

To differentiate \( \tan x \) with respect to \( \cos x \), we will use the chain rule and the concept of implicit differentiation. Here are the steps:

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = \tan x \) and \( v = \cos x \). We need to find \( \frac{du}{dv} \).

2. **Differentiate \( u \) with respect to \( x \)**:
\[ \frac{du}{dx} = \sec^2 x \]
This is because the derivative of \( \tan x \) is \( \sec^2 x \).

3. **Differentiate \( v \) with respect to \( x \)**:
\[ \frac{dv}{dx} = -\sin x \]
This is because the derivative of \( \cos x \) is \( -\sin x \).

4. **Use the chain rule to find \( \frac{du}{dv} \)**:
We know that:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]
Substituting the derivatives we found:
\[ \frac{du}{dv} = \frac{\sec^2 x}{-\sin x} \]

5. **Simplify the expression**:
\[ \frac{du}{dv} = -\frac{\sec^2 x}{\sin x} \]
We can also express \( \sec^2 x \) in terms of \( \cos x \):
\[ \sec^2 x = \frac{1}{\cos^2 x} \]
Therefore:
\[ \frac{du}{dv} = -\frac{1}{\cos^2 x \sin x} \]

### Final Answer:
\[ \frac{du}{dv} = -\frac{1}{\cos^2 x \sin x} \]