If `y=log(tanx),"then "dy/dx` is :

Options

A. `1/tanx`
B. `sec^(2)x/tanx`
C. `sec^(2)x`
D. 0

Answer

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

### Step 1: Differentiate \( y \) with respect to \( x \)
We start with the function:
\[ y = \log(\tan x) \]
To differentiate \( y \) with respect to \( x \), we apply the chain rule. The derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).

### Step 2: Identify \( u \) and find \( \frac{du}{dx} \)
Here, \( u = \tan x \). Now we need to find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \sec^2 x \]
This is because the derivative of \( \tan x \) is \( \sec^2 x \).

### Step 3: Apply the chain rule
Now we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{1}{\tan x} \cdot \frac{du}{dx} = \frac{1}{\tan x} \cdot \sec^2 x \]

### Step 4: Simplify the expression
Now we simplify the expression:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{\tan x} \]
Using the identity \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec x = \frac{1}{\cos x} \), we can rewrite \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\sec^2 x}{\frac{\sin x}{\cos x}} = \frac{\sec^2 x \cdot \cos x}{\sin x} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\sec^2 x \cdot \cos x}{\sin x} \]

### Final Result
Thus, the final result for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{\tan x} \]