Question

Use the chain rule calculate dw/dr , dw/ds and dw/dt

W=in(x+2y+3z) , x=r^2 + t^2 , y =s^2 - t^2 , z=r^2 + s^2

Answer 1

`(dw)/(dr) = (8\cdot r)/(4\cdot r^(2)-t^(2)+5\cdot s^(2))`,
`(dw)/(ds) = (10\cdot s)/(4\cdot r^(2)-t^(2)+5\cdot s^(2))`,
`(dw)/(dt) = -(2\cdot t)/(4\cdot r^(2)-t^(2)+5\cdot s^(2))`

Explanation:

We proceed to derive each expression by rule of chain. Let be
`w = \ln (x+2\cdot y + 3\cdot z)`,
`x = r^(2)+t^(2)`,
`y = s^(2)-t^(2)` and
`z = r^(2)+s^(2)`:

`(dw)/(dr) = ((dx)/(dr)+2\cdot (dy)/(dr) +3\cdot (dz)/(dr) )/(x+2\cdot y + 3\cdot z)`

`(dx)/(dr) = 2\cdot r`

`(dy)/(dr) = 0`

`(dz)/(dr) = 2\cdot r`

`(dw)/(dr) = (8\cdot r)/((r^(2)+t^(2))+2\cdot (s^(2)-t^(2))+3\cdot (r^(2)+s^(2)))`

`(dw)/(dr) = (8\cdot r)/(4\cdot r^(2)-t^(2)+5\cdot s^(2))` (1)

`(dw)/(ds) = ((dx)/(ds)+2\cdot (dy)/(ds) +3\cdot (dz)/(ds) )/(x+2\cdot y + 3\cdot z)`

`(dx)/(ds) = 0`

`(dy)/(ds) = 2\cdot s`

`(dz)/(ds) = 2\cdot s`

`(dw)/(ds) = (10\cdot s)/((r^(2)+t^(2))+2\cdot (s^(2)-t^(2))+3\cdot (r^(2)+s^(2)))`

`(dw)/(ds) = (10\cdot s)/(4\cdot r^(2)-t^(2)+5\cdot s^(2))` (2)

`(dw)/(dt) = ((dx)/(dt)+2\cdot (dy)/(dt) +3\cdot (dz)/(dt) )/(x+2\cdot y + 3\cdot z)`

`(dx)/(dt) = 2\cdot t`

`(dy)/(dt) = -2\cdot t`

`(dz)/(dt) = 0`

`(dw)/(dt) = -(2\cdot t)/((r^(2)+t^(2))+2\cdot (s^(2)-t^(2))+3\cdot (r^(2)+s^(2)))`

`(dw)/(dt) = -(2\cdot t)/(4\cdot r^(2)-t^(2)+5\cdot s^(2))` (3)