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
`(\partial w)/(\partial s)` and
`(\partial w)/(\partial t)` at the point
`(s,t)=(1,0)`.

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


`(\partial w)/(\partial s)=(\partial f)/(\partial u)(\partial u)/(\partial s)+(\partial f)/(\partial v)(\partial v)/(\partial s)`

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


`(\partial w)/(\partial s)\bigg|_((s,t)=(1,0))=(\partial f)/(\partial u)\bigg|_((u,v)=(-6,-8))(\partial u)/(\partial s)\bigg|_((s,t)=(1,0))+(\partial f)/(\partial v)\bigg|_((u,v)=(-6,-8))(\partial v)/(\partial s)\bigg|_((s,t)=(1,0))`

`(\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


`(\partial w)/(\partial t)\bigg|_((s,t)=(1,0))=(\partial f)/(\partial u)\bigg|_((u,v)=(-6,-8))(\partial u)/(\partial t)\bigg|_((s,t)=(1,0))+(\partial f)/(\partial v)\bigg|_((u,v)=(-6,-8))(\partial v)/(\partial t)\bigg|_((s,t)=(1,0))`

`(\partial w)/(\partial t)\bigg|_((s,t)=(1,0))=(-1)(7)+(2)(6)=5`