Find: `d/dx` of `( Log tan x)`

Answer

To find the derivative of \( \log(\tan x) \), we will use the chain rule and the properties of logarithmic differentiation. Here’s the step-by-step solution:

### Step 1: Identify the function
We need to differentiate the function \( y = \log(\tan x) \).

### Step 2: Apply the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) \]
This means we first differentiate the outer function (logarithm) and then multiply it by the derivative of the inner function (\(\tan x\)).

### Step 3: Differentiate \(\tan x\)
The derivative of \(\tan x\) is:
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]

### Step 4: Substitute back into the derivative
Now substituting back, we get:
\[ \frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec^2 x \]

### Step 5: Simplify the expression
We can simplify this expression further. Recall that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Thus, we can write:
\[ \frac{1}{\tan x} = \frac{\cos x}{\sin x} \]
So, substituting this into our derivative gives:
\[ \frac{dy}{dx} = \frac{\sec^2 x \cdot \cos x}{\sin x} \]
Now, since \(\sec^2 x = \frac{1}{\cos^2 x}\), we can rewrite it as:
\[ \frac{dy}{dx} = \frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = \frac{1}{\cos x \sin x} \]

### Final Answer
Thus, the derivative of \( \log(\tan x) \) is:
\[ \frac{dy}{dx} = \frac{1}{\cos x \sin x} = \frac{2}{\sin(2x)} \]