log tan x

Answer

To find the differential coefficient of the function \( f(x) = \log(\tan x) \), we will use the chain rule and the properties of logarithmic differentiation. Here is the step-by-step solution:

### Step 1: Identify the function
The function we want to differentiate is:
\[ f(x) = \log(\tan x) \]

### Step 2: Apply the chain rule
Using the chain rule, the derivative of \( \log(u) \) where \( u = \tan x \) is given by:
\[ \frac{d}{dx}[\log(u)] = \frac{1}{u} \cdot \frac{du}{dx} \]
Thus, we have:
\[ f'(x) = \frac{1}{\tan x} \cdot \frac{d}{dx}[\tan x] \]

### Step 3: Differentiate \( \tan x \)
The derivative of \( \tan x \) is:
\[ \frac{d}{dx}[\tan x] = \sec^2 x \]
Now substituting this back into our expression for \( f'(x) \):
\[ f'(x) = \frac{1}{\tan x} \cdot \sec^2 x \]

### Step 4: Simplify the expression
We can rewrite \( \sec^2 x \) in terms of sine and cosine:
\[ \sec^2 x = \frac{1}{\cos^2 x} \]
And since \( \tan x = \frac{\sin x}{\cos x} \), we can express \( f'(x) \) as:
\[ f'(x) = \frac{\sec^2 x}{\tan x} = \frac{1/\cos^2 x}{\sin x / \cos x} = \frac{1}{\sin x \cos x} \]

### Step 5: Final expression
Thus, the derivative of \( f(x) = \log(\tan x) \) is:
\[ f'(x) = \frac{1}{\sin x \cos x} = \frac{2}{\sin(2x)} \]
where we used the identity \( \sin(2x) = 2 \sin x \cos x \).

### Final Answer
The differential coefficient of \( \log(\tan x) \) is:
\[ f'(x) = \frac{2}{\sin(2x)} \]

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