The partial derivatives of the function $x=\int _u^ve^{-t^2}dt\:$ and $y=u^v$

Question

If $x=\int _u^ve^{-t^2}dt\:$ and $y=u^v$, how to find $\left(\frac{∂u}{∂x}\right)_y$, $\left(\frac{∂u}{∂y}\right)_x$, and $\left(\frac{∂y}{∂x}\right)_u$ at $u=2$ and $v=0$?

$\left(\frac{∂x}{∂u}\right)_y$ should be reciprocal with $\left(\frac{∂u}{∂x}\right)_y$, then if I take $dx=\frac{∂x}{∂v}dv+\frac{∂x}{∂u}du$

I find $\left(\frac{∂x}{∂u}\right)_y=\frac{∂}{∂u}\int _u^ve^{-t^2}dt=\frac{∂}{∂u}\int _u^ce^{-t^2}dt+\frac{∂}{∂u}\int _c^ve^{-t^2}dt=-e^{-u^2}$ thus,

$\left(\frac{∂u}{∂x}\right)_y=-e^4$ at $u=2$ and $v=0$,

I don't know which step I should take to find the values of $\left(\frac{∂u}{∂y}\right)_x$ and $\left(\frac{∂y}{∂x}\right)_u$, if you don't mind please give me some clues to solve this problem. Thank you so much.

note : it has nothing to do with homework or school.

Answer 1

finally after struggling to solve this problem, i'm arrived at the answer!

given $x=\int _u^ve^{-t^2}dt\:$ and $y=u^v$

to solve this problem, use the chain rule for $x$

$dx=\:\frac{∂x}{∂u}du+\frac{\partial x}{\partial v}dv$

solve for $\frac{∂x}{∂u}$ and $\frac{\partial x}{\partial v}$ i get

$\frac{\partial x}{\partial u}=\frac{\partial }{\partial u}\int _u^ve^{-t^2}dt=-e^{-u^2}\:\:\:\:\:\:$and$\:\:\:\:\:\:\frac{\partial x}{\partial v}=\frac{\partial }{\partial v}\int _u^ve^{-t^2}dt=e^{-v^2}$

substitute into $dx$ yields

$dx=e^{-v^2}dv-e^{-u^2}du$

now eliminate $dv$, since what we're looking for is $x$ as the function of $y$ and $u$. Since $y=u^v$, then $v=\frac{ln\left(y\right)}{ln\left(u\right)}$. Thus,

$dv=\left(\frac{1}{yln\left(u\right)}\right)dy-\left(\frac{ln\left(y\right)}{uln^2\left(u\right)}\right)du$

then substitute $dv$ into $dx$ with $y=u^v$, yields

$dx=\left(\frac{1}{e^{v^2}u^vln\left(u\right)}\right)dy-\left(e^{-u^2}+\frac{ln\left(u^v\right)}{uln^2\left(u\right)}\right)du$

at $\left\{u=2,\:v=0\right\}$,

$\left(\frac{\partial y}{\partial x}\right)_u=e^{v^2}u^vln\left(u\right)=ln\left(2\right)$

and,

$\left(\frac{\partial u}{\partial y}\right)_x=\frac{e^{u^2}}{e^{v^2}u^vln\left(u\right)}+\frac{uln\left(u\right)}{e^{v^2}u^vln\left(u^v\right)}=\frac{e^4}{ln\left(2\right)}$