`y=tanxtan2xtan3xtan4x`

Answer

To differentiate the function \( y = \tan x \tan 2x \tan 3x \tan 4x \) with respect to \( x \), we can follow these steps:

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \log y = \log(\tan x \tan 2x \tan 3x \tan 4x) \]

### Step 2: Use the property of logarithms
Using the property of logarithms that states \( \log(mn) = \log m + \log n \), we can rewrite the equation as:
\[ \log y = \log(\tan x) + \log(\tan 2x) + \log(\tan 3x) + \log(\tan 4x) \]

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

### Step 4: Apply the chain rule
Using the chain rule, we know that:
\[ \frac{d}{dx}[\log(\tan kx)] = \frac{1}{\tan kx} \cdot \sec^2(kx) \cdot k \]
for \( k = 1, 2, 3, 4 \).

Thus, we can write:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{\tan x} \sec^2 x + \frac{1}{\tan 2x} \cdot 2 \sec^2(2x) + \frac{1}{\tan 3x} \cdot 3 \sec^2(3x) + \frac{1}{\tan 4x} \cdot 4 \sec^2(4x) \]

### Step 5: Multiply through by \( y \)
Now, we multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( \frac{\sec^2 x}{\tan x} + \frac{2 \sec^2(2x)}{\tan 2x} + \frac{3 \sec^2(3x)}{\tan 3x} + \frac{4 \sec^2(4x)}{\tan 4x} \right) \]

### Step 6: Substitute back for \( y \)
Since \( y = \tan x \tan 2x \tan 3x \tan 4x \), we can substitute back:
\[ \frac{dy}{dx} = \tan x \tan 2x \tan 3x \tan 4x \left( \frac{\sec^2 x}{\tan x} + \frac{2 \sec^2(2x)}{\tan 2x} + \frac{3 \sec^2(3x)}{\tan 3x} + \frac{4 \sec^2(4x)}{\tan 4x} \right) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \tan x \tan 2x \tan 3x \tan 4x \left( \frac{\sec^2 x}{\tan x} + \frac{2 \sec^2(2x)}{\tan 2x} + \frac{3 \sec^2(3x)}{\tan 3x} + \frac{4 \sec^2(4x)}{\tan 4x} \right) \]