The differential coefficient of log tan x is

Options

A. `2 sec 2 x `
B. ` 2 "cosec" 2 x`
C. `2 sec^3 x`
D. `2 " cosec"^3x`

Answer

To find the differential coefficient of \( \log(\tan x) \), we will apply the chain rule of differentiation. Here’s the step-by-step solution:

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

### Step 2: Apply the chain rule
Using the chain rule, we differentiate the outer function (logarithm) first:
\[ \frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{d}{dx}(\tan x) \]

### Step 3: Differentiate the inner function
Now, we need to differentiate \( \tan x \):
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]

### Step 4: Substitute back into the equation
Substituting this back into our equation gives:
\[ \frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec^2 x \]

### Step 5: Simplify the expression
We can simplify this expression:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{\tan x} \]

### Step 6: Rewrite in terms of sine and cosine
Recall that:
\[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \sec x = \frac{1}{\cos x} \]
Thus:
\[ \sec^2 x = \frac{1}{\cos^2 x} \]
So we can rewrite \( \frac{\sec^2 x}{\tan x} \) as:
\[ \frac{\sec^2 x}{\tan x} = \frac{\frac{1}{\cos^2 x}}{\frac{\sin x}{\cos x}} = \frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = \frac{1}{\sin x \cos x} \]

### Step 7: Final expression
Thus, the final expression for the differential coefficient of \( \log(\tan x) \) is:
\[ \frac{dy}{dx} = \frac{1}{\sin x \cos x} \]