If u=e^(xyz),show that d^(3)u/dx*dy*dz=(1+ 3xyz+x^(2) y^(2) z^(2) )*e^(xyz)​

To show that \( \frac{d^3u}{dx \, dy \, dz} = (1 + 3xyz + x^2 y^2 z^2) e^{xyz} \) for the function \( u = e^{xyz} \), we will need to compute the third mixed partial derivative of \( u \) with respect to \( x \), \( y \), and \( z \). Let's break this down step by step.

Step 1: First Partial Derivative

We start by finding the first partial derivative of \( u \) with respect to \( x \):

Using the chain rule, we have:

\[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^{xyz}) = e^{xyz} \cdot \frac{\partial}{\partial x}(xyz) = e^{xyz} \cdot (yz) \]

Thus, the first derivative is:

\[ \frac{\partial u}{\partial x} = yze^{xyz} \]

Step 2: Second Partial Derivative

Next, we differentiate \( \frac{\partial u}{\partial x} \) with respect to \( y \):

Applying the product rule:

\[ \frac{\partial^2 u}{\partial y \partial x} = \frac{\partial}{\partial y}(yze^{xyz}) = z e^{xyz} + yz \cdot \frac{\partial}{\partial y}(e^{xyz}) = z e^{xyz} + yz \cdot (xze^{xyz}) = z e^{xyz} + xyz^2 e^{xyz} \]

Combining terms gives us:

\[ \frac{\partial^2 u}{\partial y \partial x} = (z + xyz^2)e^{xyz} \]

Step 3: Third Partial Derivative

Now, we differentiate \( \frac{\partial^2 u}{\partial y \partial x} \) with respect to \( z \):

Again, using the product rule:

\[ \frac{\partial^3 u}{\partial z \partial y \partial x} = \frac{\partial}{\partial z}((z + xyz^2)e^{xyz}) = (1 + 2xyz)e^{xyz} + (z + xyz^2)(xy e^{xyz}) \]

Breaking this down:

• The first term is \( (1 + 2xyz)e^{xyz} \).
• The second term expands to \( (z + xyz^2)(xy)e^{xyz} = xyze^{xyz} + x^2y^2z^2e^{xyz} \).

Combining these results, we have:

\[ \frac{\partial^3 u}{\partial x \partial y \partial z} = (1 + 3xyz + x^2y^2z^2)e^{xyz} \]

Final Result

Thus, we have shown that:

\[ \frac{d^3u}{dx \, dy \, dz} = (1 + 3xyz + x^2y^2z^2)e^{xyz} \]

This result illustrates how the mixed partial derivatives of an exponential function can be computed systematically using the product and chain rules. Each step builds on the previous derivatives, showcasing the interconnectedness of calculus concepts.