`y=tanxtan2xtan3xtan4x`

Answer

To solve the problem, we need to find the derivative of the function \( y = \tan x \tan 2x \tan 3x \tan 4x \). We can use logarithmic differentiation to simplify the process. Here’s a step-by-step solution:

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

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

### 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, the derivative of \( \log(\tan kx) \) is:
\[ \frac{d}{dx}(\log(\tan kx)) = \frac{1}{\tan kx} \cdot \sec^2(kx) \cdot k \]
Thus, we have:
\[ \frac{d}{dx}(\log(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x \cdot 1 \]
\[ \frac{d}{dx}(\log(\tan 2x)) = \frac{1}{\tan 2x} \cdot \sec^2(2x) \cdot 2 \]
\[ \frac{d}{dx}(\log(\tan 3x)) = \frac{1}{\tan 3x} \cdot \sec^2(3x) \cdot 3 \]
\[ \frac{d}{dx}(\log(\tan 4x)) = \frac{1}{\tan 4x} \cdot \sec^2(4x) \cdot 4 \]

### Step 5: Combine the derivatives
Now we can combine these results:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{\sec^2 x}{\tan x} + \frac{2 \sec^2(2x)}{\tan 2x} + \frac{3 \sec^2(3x)}{\tan 3x} + \frac{4 \sec^2(4x)}{\tan 4x} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \frac{\sec^2 x}{\tan x} + \frac{2 \sec^2(2x)}{\tan 2x} + \frac{3 \sec^2(3x)}{\tan 3x} + \frac{4 \sec^2(4x)}{\tan 4x} \right) \]

### Step 7: Substitute back for \( y \)
Substituting \( y = \tan x \tan 2x \tan 3x \tan 4x \):
\[ \frac{dy}{dx} = \tan x \tan 2x \tan 3x \tan 4x \left( \frac{\sec^2 x}{\tan x} + \frac{2 \sec^2(2x)}{\tan 2x} + \frac{3 \sec^2(3x)}{\tan 3x} + \frac{4 \sec^2(4x)}{\tan 4x} \right) \]

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