Question

Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8)=2

Answer 1

$w(s,t)=f(u(s,t),v(s,t))$

From the given set of conditions, it's likely that you are asked to find the values of $\dfrac{\partial w}{\partial s}$ and $\dfrac{\partial w}{\partial t}$ at the point $(s,t)=(1,0)$.

By the chain rule, the partial derivative with respect to $s$ is

$\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}$

and so at the point $(1,0)$, we have

$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21$

Similarly, the partial derivative with respect to $t$ would be found via

$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5$