Website title: Partial Differentiation for f(x,y)=(3x^2+y^2+2xy)^1/2

[wezzo62]

Find ∂f_xand ∂f_y for: f(x,y)=(3x²+y²+2xy)^1/2

i tried using the chain rule and said f(x,y) = u^1/2 then
∂f_u = 1/2u^-1/2, ∂u_x = 6x + 2 and ∂u_y=2y+2

∂f_x = 1/2(3x² + y² + 2xy)^-1/2(6x + 2)

∂f_y = 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 ∂u_x and ∂u_y part right?

I also have a similar problem with f(x,y)=(x+y)³:
∂f_u=3u², ∂u_x and ∂u_y=1
in which case both ∂f_x and ∂f_y=3(x+y)². . . .
im not sure if ∂u_x should be 1+y, ∂u_y = x+1 therefore
∂f_x=3(x+y)²(1+y) and ∂f_y=3(x+y)²(x+1)

any help much appreciated, thanks

 

[Char. Limit]

Your first one is almost correct, and your second has the same error. Remember that the derivative of 2xy with respect to y is 2x, not 2. And similarly with the derivative with respect to x: it's 2y, not 2.

 

[wezzo62]

I figured it was something like that. so for the first one ∂u_x=6x+y²+2y and ∂u_y=3x²+2y+2x? which would make ∂f_x=1/2(3x²+y²+2xy)^-1/2(6x+y²+2y) and ∂f_y=1/2(3x²+y²+2xy)^-1/2(3x²+2y+2x) . . . . ?

also for the second one I've now got ∂f_x=3(x+y)²y and ∂f_y=3(x+y)²x

 

[Char. Limit]

You've almost got it, but now you're including too many terms. What's the derivative of y^2 with respect to x?

 

[wezzo62]

2y dy/dx?

 

[Char. Limit]

That would be correct if y depended on x, but here y and x are independent. So y^2 is a constant with respect to x, and you know the derivative of a constant, yeah?