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 follow these steps:

### Step 1: Define the function
Let \( y = \log(\tan x) \).

### Step 2: Apply the chain rule
To differentiate \( y \), we will use the chain rule. The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

### Step 3: Differentiate the outer function
The outer function is \( \log(u) \), where \( u = \tan x \). The derivative of \( \log(u) \) is:
\[ \frac{d}{du}(\log(u)) = \frac{1}{u} \]
So, applying this to our function:
\[ \frac{dy}{du} = \frac{1}{\tan x} \]

### Step 4: Differentiate the inner function
Now we need to differentiate the inner function \( u = \tan x \). The derivative of \( \tan x \) is:
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]

### Step 5: Combine the derivatives
Now, we can combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\tan x} \cdot \sec^2 x \]

### Step 6: Simplify the expression
We know that \( \sec x = \frac{1}{\cos x} \) and \( \tan x = \frac{\sin x}{\cos x} \). Therefore:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{\tan x} = \frac{\sec^2 x}{\frac{\sin x}{\cos x}} = \frac{\sec^2 x \cdot \cos x}{\sin x} \]
Since \( \sec^2 x = \frac{1}{\cos^2 x} \), we can write:
\[ \frac{dy}{dx} = \frac{1}{\cos^2 x} \cdot \cos x \cdot \frac{1}{\sin x} = \frac{1}{\cos x \cdot \sin x} \]

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