If u = log(x + y + z − 3xyz) show that (i) u + u + u =

xyz

3

x+y+z (ii) uxx + uyy + uzz =

−

3

(x+y+z)2

(iii) ( ∂

∂x

+

∂

∂y

+

∂

∂z

2

) u=−

(x+y+z)2 .

9

Given the function \( u = \log(x + y + z - 3xyz) \), we need to show the following three results:

### (i) \( u_x + u_y + u_z = \frac{xyz}{x + y + z} \)

First, let's find the partial derivatives of \( u \) with respect to \( x \), \( y \), and \( z \). 

#### Step 1: Compute \( u_x \)

The partial derivative of \( u \) with respect to \( x \) is:

\[
u_x = \frac{\partial}{\partial x} \log(x + y + z - 3xyz)
\]

Using the chain rule:

\[
u_x = \frac{1}{x + y + z - 3xyz} \cdot \frac{\partial}{\partial x} (x + y + z - 3xyz)
\]

Now, calculate the partial derivative of the argument of the logarithm:

\[
\frac{\partial}{\partial x} (x + y + z - 3xyz) = 1 - 3yz
\]

Thus:

\[
u_x = \frac{1 - 3yz}{x + y + z - 3xyz}
\]

#### Step 2: Compute \( u_y \)

By symmetry, the partial derivative of \( u \) with respect to \( y \) is:

\[
u_y = \frac{1 - 3xz}{x + y + z - 3xyz}
\]

#### Step 3: Compute \( u

If u = log(x + y + z − 3xyz), show that (i) ux + uy + uz = xyz / (x + y + z) (ii) uxx + uyy + uzz = -3 / (x + y + z)^2 (iii) (∂/∂x + ∂/∂y + ∂/∂z)^2 u = -9 / (x + y + z)^2.

Explore how to compute partial derivatives for the function u = log(x + y + z − 3xyz). This solution demonstrates key multivariable calculus concepts, including partial derivatives and logarithmic differentiation, with step-by-step solutions for problems involving second derivatives and combined partials.

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