If `y = tan x tan 2x tan 3x, (sin 12x != 0)` then `dy / dx` has the value equal to

Options

A. `3sec^(2)3xtanx tan2x+sec^(2)x tan2xtan3x+2sec^(2)2xtan3xtanx`
B. `2y(cosec 2x+2 cosec 4x+3cosec6x)`
C. `3sec^(2)3x-2sec^(2)2x-sec^(2)x`
D. `sec^(2)x+2sec^(2)2x+3sec^(2)3x`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \tan x \tan 2x \tan 3x \), we can follow these steps:

### Step 1: Write the function
We start with the function:
\[ y = \tan x \tan 2x \tan 3x \]

### Step 2: Take the logarithm of both sides
To simplify differentiation, we take the natural logarithm of both sides:
\[ \log y = \log(\tan x \tan 2x \tan 3x) \]

### Step 3: Use the logarithmic property
Using the property of logarithms, we can express the right side as a sum:
\[ \log y = \log(\tan x) + \log(\tan 2x) + \log(\tan 3x) \]

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec^2 x + \frac{2}{\tan 2x} \cdot \sec^2 2x + \frac{3}{\tan 3x} \cdot \sec^2 3x \]

### Step 5: Multiply by \( y \)
Now, we multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( \frac{1}{\tan x} \sec^2 x + \frac{2}{\tan 2x} \sec^2 2x + \frac{3}{\tan 3x} \sec^2 3x \right) \]

### Step 6: Substitute back for \( y \)
Substituting back \( y = \tan x \tan 2x \tan 3x \):
\[ \frac{dy}{dx} = \tan x \tan 2x \tan 3x \left( \frac{1}{\tan x} \sec^2 x + \frac{2}{\tan 2x} \sec^2 2x + \frac{3}{\tan 3x} \sec^2 3x \right) \]

### Step 7: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = \tan 2x \tan 3x \sec^2 x + 2 \tan x \tan 3x \sec^2 2x + 3 \tan x \tan 2x \sec^2 3x \]

### Final expression
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \tan 2x \tan 3x \sec^2 x + 2 \tan x \tan 3x \sec^2 2x + 3 \tan x \tan 2x \sec^2 3x \]