I figured it was something like that. so for the first one ∂ux=6x+y2+2y and ∂uy=3x2+2y+2x? which would make ∂fx=1/2(3x2+y2+2xy)-1/2(6x+y2+2y) and ∂fy=1/2(3x2+y2+2xy)-1/2(3x2+2y+2x) . . . . ?
also for the second one I've now got ∂fx=3(x+y)2y and ∂fy=3(x+y)2x
Find ∂fxand ∂fy for: f(x,y)=(3x2+y2+2xy)1/2
i tried using the chain rule and said f(x,y) = u1/2 then
∂fu = 1/2u-1/2, ∂ux = 6x + 2 and ∂uy=2y+2
∂fx = 1/2(3x2 + y2 + 2xy)-1/2(6x + 2)
∂fy = 1/2(3x2 + y2 + 2xy)-1/2(2y + 2)
im not sure if this is correct as i don't know if I am doing the ∂ux and...
Find the general solution to: dy/dx = 3x2y
I tried saying u = 3x2 and v = y
then du/dv = 6x and dv/dy = 1
and get 3x2 + 6xy but now i think iv gone completely the wrong way around this . . . .
Find the general solution to:
dy/dx = ex+y
Im not sure if I am doing this right or not. i tried saying ex+y = ex x ey and using partial differentiation to solve it but keep getting the same as the question: ex+y
i know differentiating ex gives ex so is it the same in this case?