Differentiate (log x) with repect to tan x.

Answer

To differentiate \( \log x \) with respect to \( \tan x \), we will follow these steps:

### Step 1: Define the functions
Let:
- \( y = \log x \)
- \( t = \tan x \)

### Step 2: Use the chain rule
We want to find \( \frac{dy}{dt} \). By using the chain rule, we can express this as:
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]
This is our equation (1).

### Step 3: Differentiate \( y = \log x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{x} \]
This is our equation (2).

### Step 4: Differentiate \( t = \tan x \)
Next, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \sec^2 x \]
From this, we can find \( \frac{dx}{dt} \) by taking the reciprocal:
\[ \frac{dx}{dt} = \frac{1}{\sec^2 x} \]
This is our equation (3).

### Step 5: Substitute into the chain rule equation
Now we substitute equations (2) and (3) into equation (1):
\[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \left(\frac{1}{x}\right) \cdot \left(\frac{1}{\sec^2 x}\right) \]

### Step 6: Simplify the expression
This simplifies to:
\[ \frac{dy}{dt} = \frac{1}{x \sec^2 x} \]
We can also express \( \sec^2 x \) in terms of cosine:
\[ \sec^2 x = \frac{1}{\cos^2 x} \]
Thus, we can rewrite the expression as:
\[ \frac{dy}{dt} = \frac{\cos^2 x}{x} \]

### Final Answer
The derivative of \( \log x \) with respect to \( \tan x \) is:
\[ \frac{dy}{dt} = \frac{1}{x \sec^2 x} \quad \text{or} \quad \frac{\cos^2 x}{x} \]

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