I want to check that I understand this correctly before heading off to the exam.
If $z = \sin(xe^y)$ where $x = 3u^2 + uv$ and $y = u^3 - \ln(v)$ find $\displaystyle \frac{\partial z}{\partial u}$
I have this:
$\displaystyle \frac{\partial z}{\partial u} = \displaystyle \frac{\partial z}{\partial x} \cdot \displaystyle \frac{\partial x}{\partial u}$
$\displaystyle \frac{\partial z}{\partial x} \sin(xe^y) = x^y\cos(xe^y)$
$\displaystyle \frac{\partial x}{\partial u} 3u^2 + uv = 6u + v$
\begin{align} \displaystyle \frac{\partial z}{\partial u} &= \displaystyle \frac{\partial z}{\partial x} \cdot \displaystyle \frac{\partial x}{\partial u}\\ &= x^y\cos(xe^y) .(6u + v)\\ &= (3u^2 + uv)^{u^3 - \ln(v)}\cos((3u^2 + uv)e^{u^3 - \ln(v)}) .(6u + v) \end{align}
This still needs some simplification, but is this the correct approach?
I have noticed that this question is still unanswered. $$ \partial_u z = \partial_x z \partial_u x + \partial_y z \partial_u y $$ where $$ z = \sin\left(x\mathrm{e}^y\right),\\ x = 3u^2 + uv,\\ y = u^3 - \ln(v). $$
this leads to $$ \partial_x z = \mathrm{e}^y\cos(x\mathrm{e}^y),\\ \partial_y z = x\mathrm{e}^{y}\cos(x\mathrm{e}^y). $$ for the partials w.r.t u of the variables x and y we use implicit differentiation. $$ \partial_u x = 6u + v,\\ \partial_u y = 6u^2. $$
Then $$ \partial_u z = \mathrm{e}^y\cos(x\mathrm{e}^y)*(6u + v) + x\mathrm{e}^{y}\cos(x\mathrm{e}^y)*(6u^2)\\ =\mathrm{e}^y\cos(x\mathrm{e}^y)\left[6u + v + x6u^2\right] $$ Now i don't like to leave the result in terms of the original variables, so $$ \partial_u z = \dfrac{\mathrm{e}^{u^3}}{v}\sqrt{1-z^2}\left[6u + v + 6(3u^2+uv)u^2\right]\\ = \dfrac{\mathrm{e}^{u^3}}{v}\sqrt{1-z^2}\left[6u\left(1 + 3u^3\right) + v\left(1+6u^3\right)\right] $$
p.s I hope your exam went well :).