Answer
${\frac {\partial u}{\partial x}} = y^{z}$
${\frac {\partial u}{\partial y}} = x*y^{z-1}*z$
${\frac {\partial u}{\partial z}} = x*y^{z}*\ln(y)$
${\frac {\partial u}{\partial x}}(\frac{\pi}{4} ; \frac{1}{2} ; \frac{-1}{2}) = \sqrt {2}$
${\frac {\partial u}{\partial y}}(\frac{\pi}{4} ; \frac{1}{2} ; \frac{-1}{2}) = \frac{-\pi*\sqrt {2}}{4}$
${\frac {\partial u}{\partial z}}(\frac{\pi}{4} ; \frac{1}{2} ; \frac{-1}{2}) = \frac{\pi*\sqrt {2}*\ln(\frac{1}{2})}{4}$
Work Step by Step
a) We express ${\frac {\partial u}{\partial x}}$,${\frac {\partial u}{\partial y}}$, and ${\frac {\partial u}{\partial z}}$ in the following ways:
1) as functions of $x$,$y$ and $z$ by using the Chain Rule:
${\frac {\partial u}{\partial x}}$ = ${\frac {\partial u}{\partial p}}$*$\frac{dp}{dx}$ + ${\frac {\partial u}{\partial q}}$*$\frac{dq}{dx}$ + ${\frac {\partial u}{\partial r}}$*$\frac{dr}{dx}$ = ${\frac {\partial }{\partial p}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dx}[sin(x)]$ + ${\frac {\partial }{\partial q}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dx}[z^{2}*\ln(y)]$ + ${\frac {\partial }{\partial r}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dx}[\frac{1}{z}]$ = $\frac{e^{qr}}{\sqrt {1-p^{2}}}$
where $p = sin(x)$, $q = z^{2}*\ln(y)$ and $r = {\frac {1}{z}}$
${\frac {\partial u}{\partial x}}$ = $\frac{e^{z^{2}*\ln(y)*\frac{1}{z}}}{\sqrt {1 - sin^{2}(x)}}$ = $y^{z}$
${\frac {\partial u}{\partial y}}$ = ${\frac {\partial u}{\partial p}}$*$\frac{dp}{dy}$ + ${\frac {\partial u}{\partial q}}$*$\frac{dq}{dy}$ + ${\frac {\partial u}{\partial r}}$*$\frac{dr}{dy}$ = ${\frac {\partial }{\partial p}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dy}[sin(x)]$ + ${\frac {\partial }{\partial q}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dy}[z^{2}*\ln(y)]$ + ${\frac {\partial }{\partial r}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dy}[\frac{1}{z}]$ = $sin^{-1}(p)*e^{qr}*r*\frac{z^{2}}{y}$
where $p = sin(x)$, $q = z^{2}*\ln(y)$ and $r = {\frac {1}{z}}$
${\frac {\partial u}{\partial x}}$ = $sin^{-1}(sin(x))*e^{z^{2}*\ln(y)*\frac{1}{z}}*\frac{1}{z}*\frac{z^{2}}{y}$ = $x*y^{z-1}*z$
${\frac {\partial u}{\partial z}}$ = ${\frac {\partial u}{\partial p}}$*$\frac{dp}{dz}$ + ${\frac {\partial u}{\partial q}}$*$\frac{dq}{dz}$ + ${\frac {\partial u}{\partial r}}$*$\frac{dr}{dz}$ = ${\frac {\partial }{\partial p}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dz}[sin(x)]$ + ${\frac {\partial }{\partial q}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dz}[z^{2}*\ln(y)]$ + ${\frac {\partial }{\partial r}}[e^{q*r}*sin^{-1}(p)]$*$\frac{d}{dz}[\frac{1}{z}]$ = $sin^{-1}(p)*e^{qr}*(2z*\ln(y)*r-\frac{q}{z^{2}})$
where $p = sin(x)$, $q = z^{2}*\ln(y)$ and $r = {\frac {1}{z}}$
${\frac {\partial u}{\partial x}}$ = $sin^{-1}(sin(x))*e^{z^{2}*\ln(y)*\frac{1}{z}}*(2*z*\ln(y)*\frac{1}{z} - \frac{z^{2}*\ln(y)}{z^{2}})$ = $x*y^{z}*\ln(y)$
2) by expressing $u$ directly in terms of $x$,$y$, and $z$ before differentiating:
$u = e^{q*r}*sin^{-1}(p)$
where $p = sin(x)$, $q = z^{2}*\ln(y)$ and $r = {\frac {1}{z}}$
$u = e^{z^{2}*\ln(y)*\frac {1}{z}}*sin^{-1}(sin(x)) = y^{z}*x$
${\frac {\partial u}{\partial x}}$ = ${\frac {\partial }{\partial x}}[y^{z}*x] = y^{z}$
${\frac {\partial u}{\partial y}}$ = ${\frac {\partial }{\partial y}}[y^{z}*x] = x*y^{z-1}*z$
${\frac {\partial u}{\partial z}}$ = ${\frac {\partial }{\partial z}}[y^{z}*x] = x*y^{z}*\ln(y)$
b) Evaluate ${\frac {\partial u}{\partial x}}$,${\frac {\partial u}{\partial y}}$, and ${\frac {\partial u}{\partial z}}$ at the given point($x$,$y$,$z$):
($x$,$y$,$z$)=($\frac{\pi}{4}$ ; $\frac{1}{2}$ ; $\frac{-1}{2}$)
Thus:
${\frac {\partial u}{\partial x}}(\frac{\pi}{4} ; \frac{1}{2} ; \frac{-1}{2}) = \sqrt {2}$
${\frac {\partial u}{\partial y}}(\frac{\pi}{4} ; \frac{1}{2} ; \frac{-1}{2}) = \frac{-\pi*\sqrt {2}}{4}$
${\frac {\partial u}{\partial z}}(\frac{\pi}{4} ; \frac{1}{2} ; \frac{-1}{2}) = \frac{\pi*\sqrt {2}*\ln(\frac{1}{2})}{4}$
