Partial Derivatives 1

Notice that $\dfrac{\partial}{\partial z} \left(\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)\right)$ will only leave terms with $z$ , since $\dfrac{\partial}{\partial z}$ of all other terms with $x$ and $y$ will go to $0$ . Also notice that $\dfrac{\partial}{\partial x} \big(- \ln (z + 1) ^{\pi}\big) = 0$ . Thus,

$\begin{aligned} \dfrac{\partial}{\partial z} \left(\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)\right) & = \dfrac{\partial}{\partial z} \left(\dfrac{\partial}{\partial y}\left(\dfrac{\partial}{\partial x}\left(e^{xyz}\right)\right)\right) \\ & = \dfrac{\partial}{\partial z} \left(\dfrac{\partial}{\partial y}\left(yze^{xyz}\right)\right) \\ & = \dfrac{\partial}{\partial z} \left(ze^{xyz} + xyz^2e^{xyz}\right) \\ & = e^{xyz} \left(1 + 3xyz + (xyz)^2\right) \end{aligned}$

Plugging in our point $(x,y,z) = (2,-4,1)$ , we have $e^{-8}\left(1 - 24 + 64\right) = 41e^{-8}$ , so our answer is $\boxed{41}$ .

$\begin{aligned} f(x,y,z) & = x^{3y}\cos (y \ln x) + e^{xyz} - \color{#3D99F6} \ln (z+1)^\pi & \small \color{#3D99F6} \text{Term is independent of }x. \\ \frac {\partial f}{\partial x} & = \frac {\partial}{\partial x} x^{3y}\cos (y \ln x) + \frac {\partial}{\partial x} e^{xyz} - \color{#3D99F6} 0 \\ \frac {\partial}{\partial y} \left(\frac {\partial f}{\partial x}\right) & = {\color{#3D99F6} \frac {\partial}{\partial y} \left(\frac {\partial}{\partial x} x^{3y}\cos (y \ln x)\right)} + \frac {\partial}{\partial y} \left(\frac {\partial}{\partial x} e^{xyz}\right) & \small \color{#3D99F6} \text{Term is independent of }z. \\ \frac {\partial}{\partial z} \left(\frac {\partial}{\partial y} \left(\frac {\partial f}{\partial x}\right) \right) & = {\color{#3D99F6} 0} + \frac {\partial}{\partial z} \left(\frac {\partial}{\partial y} \left(\frac {\partial}{\partial x} e^{xyz}\right) \right) \\ & = \frac {\partial}{\partial z} \left(\frac {\partial}{\partial y} yze^{xyz}\right) \\ & = \frac {\partial}{\partial z} \left(ze^{xyz} + xyz^2e^{xyz} \right) \\ & = e^{xyz} + xyze^{xyz} + 2xyze^{xyz} + x^2y^2z^2e^{xyz} \\ & = e^{xyz} \left(1+ 3xyz + x^2y^2z^2\right) \end{aligned}$

Therefore,

$\begin{aligned} \frac {\partial}{\partial z} \left(\frac {\partial}{\partial y} \left(\frac {\partial f}{\partial x}\right) \right) \bigg|_{(2,-4,1)} & = e^{-8} \left(1- 24 + 64 \right) = \frac {41}{e^8}\end{aligned}$

$\implies n = \boxed{41}$