Q14Mathematics I

Question

If $u = e^{x y z}$ , then show that:

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

Answer 1

The mathematical objective is to rigorously calculate the third-order mixed partial derivative $\frac{\partial ^{3} u}{\partial x \partial y \partial z}$ (or $u_{x y z}$ ) of the exponential function $u = e^{x y z}$ . By Clairaut's Theorem, for well-behaved continuous functions, the order of partial differentiation does not matter. We will proceed systematically by differentiating with respect to $x$ , then $y$ , and finally $z$ .

Step 1: First Derivative w.r.t $x$

We differentiate $u$ partially with respect to $x$ . During this operation, both $y$ and $z$ are treated strictly as mathematical constants. We utilize the standard chain rule:

u_{x} = \frac{\partial}{\partial x} (e^{x y z}) = e^{x y z} \cdot \frac{\partial}{\partial x} (x y z) = y z e^{x y z}

Step 2: Second Derivative w.r.t $y$

Next, we differentiate the result $u_{x}$ with respect to $y$ , holding $x$ and $z$ as strict constants. Because the variable $y$ appears in both the polynomial coefficient $(y z)$ and the exponential exponent $(x y z)$ , we must meticulously apply the Product Rule: $\frac{\partial}{\partial y} (A \cdot B) = A^{'} B + A B^{'}$ .

u_{x y} = \frac{\partial}{\partial y} (y z \cdot e^{x y z})

u_{x y} = (\frac{\partial}{\partial y} (y z)) e^{x y z} + (y z) (\frac{\partial}{\partial y} e^{x y z})

u_{x y} = (z) e^{x y z} + (y z) (x z e^{x y z}) = (z + x y z^{2}) e^{x y z}

Step 3: Third Derivative w.r.t $z$

Finally, we differentiate $u_{x y}$ strictly with respect to $z$ , treating $x$ and $y$ as constants. Again, because $z$ appears in both the grouped polynomial term and the exponential term, the Product Rule is absolutely required.

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

u_{x y z} = (\frac{\partial}{\partial z} (z + x y z^{2})) e^{x y z} + (z + x y z^{2}) (\frac{\partial}{\partial z} e^{x y z})

Executing the strict partial derivatives yields:

u_{x y z} = (1 + 2 x y z) e^{x y z} + (z + x y z^{2}) (x y e^{x y z})

We now elegantly factor out the common $e^{x y z}$ term and algebraically expand the brackets:

u_{x y z} = e^{x y z} [(1 + 2 x y z) + x y (z + x y z^{2})]

u_{x y z} = e^{x y z} (1 + 2 x y z + x y z + x^{2} y^{2} z^{2})

Combining the like terms mathematically finalizes the rigorous proof:

u_{x y z} = e^{x y z} (1 + 3 x y z + x^{2} y^{2} z^{2})

Q14Mathematics I

Question

Answer

Step 1: First Derivative w.r.t x

Step 2: Second Derivative w.r.t y

Step 3: Third Derivative w.r.t z

Step 1: First Derivative w.r.t $x$

Step 2: Second Derivative w.r.t $y$

Step 3: Third Derivative w.r.t $z$