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The problem given is:

If \( u = \log(\tan x + \tan y + \tan z) \), show that

\[
\sin 2x \frac{\partial u}{\partial x} + \sin 2y \frac{\partial u}{\partial y} + \sin 2z \frac{\partial u}{\partial z} = 2.
\]

### Solution Outline:

1. First, compute the partial derivatives \( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z} \).

Given \( u = \log(\tan x + \tan y + \tan z) \), we can differentiate using the chain rule.

2. Use the fact that the derivative of \( \log(f(x)) \) is \( \frac{1}{f(x)} \cdot f'(x) \), to handle the logarithmic function.

3. Next, apply the trigonometric identities to simplify the expressions involving \( \sin 2x, \sin 2y, \sin 2z \).

4. Combine the results and simplify to show the desired result.

Let me proceed step by step.

### Step 1: Compute Partial Derivatives

\[
\frac{\partial u}{\partial x} = \frac{1}{\tan x + \tan y + \tan z} \cdot \frac{d}{dx} (\tan x) = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2 x
\]

\[
\frac{\partial u}{\partial y} = \frac{1}{\tan x + \tan y + \tan z} \cdot \frac{d}{dy} (\tan y) = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2 y
\]

\[
\frac{\partial u}{\partial z} = \frac{1}{\tan x + \tan y + \tan z} \cdot \frac{d}{dz} (\tan z) = \frac{1}{\tan x + \tan y + \tan z} \cdot \sec^2 z
\]

### Step 2: Multiply by Sine Terms

Now, let's multiply these partial derivatives by the respective sine terms:

\[
\sin 2x \frac{\partial u}{\partial x} = \sin 2x \cdot \frac{\sec^2 x}{\tan x + \tan y + \tan z}
\]

\[
\sin 2y \frac{\partial u}{\partial y} = \sin 2y \cdot \frac{\sec^2 y}{\tan x + \tan y + \tan z}
\]

\[
\sin 2z \frac{\partial u}{\partial z} = \sin 2z \cdot \frac{\sec^2 z}{\tan x + \tan y + \tan z}
\]

### Step 3: Trigonometric Identity

Recall that \( \sin 2x = 2 \sin x \cos x \), and that \( \sec^2 x = 1 + \tan^2 x \). This simplifies the terms as follows:

\[
\sin 2x \cdot \sec^2 x = 2 \sin x \cos x \cdot (1 + \tan^2 x) = 2 \tan x
\]

Similarly, for the other terms:

\[
\sin 2y \cdot \sec^2 y = 2 \tan y
\]
\[
\sin 2z \cdot \sec^2 z = 2 \tan z
\]

### Step 4: Add the Terms

Now, adding these results together:

\[
\frac{2 \tan x}{\tan x + \tan y + \tan z} + \frac{2 \tan y}{\tan x + \tan y + \tan z} + \frac{2 \tan z}{\tan x + \tan y + \tan z}
\]

Since the numerators are all multiplied by 2, and the denominators are the same, the terms combine to give:

\[
2 \frac{\tan x + \tan y + \tan z}{\tan x + \tan y + \tan z} = 2
\]

Thus, we have shown that:

\[
\sin 2x \frac{\partial u}{\partial x} + \sin 2y \frac{\partial u}{\partial y} + \sin 2z \frac{\partial u}{\partial z} = 2
\]

Would you like any details or have further questions?

Here are 5 related questions for you:
1. How does the chain rule apply to logarithmic functions?
2. What is the significance of trigonometric identities in simplifying expressions?
3. Can you generalize this result for other logarithmic functions?
4. How would the result change if the argument of the logarithm were different?
5. How do partial derivatives differ from total derivatives in multivariable functions?

**Tip:** When differentiating logarithmic functions, remember that the chain rule can significantly simplify the process.

If u = log(tan x + tan y + tan z), show that sin 2x ∂u/∂x + sin 2y ∂u/∂y + sin 2z ∂u/∂z = 2.

Learn how to compute partial derivatives and apply trigonometric identities to prove that sin 2x ∂u/∂x + sin 2y ∂u/∂y + sin 2z ∂u/∂z = 2. This solution involves using the chain rule, logarithmic differentiation, and multivariable calculus.

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