How do you differentiate f(2x-y^2) wrt to x and y?
I would like a method or link to a webpage explaining how to do it.
It's probably chain rule, but I don't know how to apply it here.

Question

How do you differentiate f(2x-y^2) wrt to x and y?
I would like a method or link to a webpage explaining how to do it.
It's probably chain rule, but I don't know how to apply it here.

anonymous · Answer

single variable or multivariable? and is it f(x,y) = 2x-y^2?

henryblah · Answer

The whole problem is u(x,y)=(e^-x)f(2x-y^2). And I need to find partial differentiation du/dx and dv/dx. So I think it's multivariable but not sure.

anonymous · Answer

yes, that's multivariable.

anonymous · Answer

what is f(x) or f(x,y) defined as?

henryblah · Answer

umm, it's a function from R to R and is differentiable.

henryblah · Answer

I have to prove it satisfies a u(x,y) satisfies a differential equation if that helps.

anonymous · Answer

I haven't done anything with differential equations in multiple dimensions yet, sorry

henryblah · Answer

No worries

henryblah · Answer

Here's the question if anyone is interested http://i.imgur.com/7QTnYN1.jpg

anonymous · Answer

f(x)'=2
f(y)'=-2y

henryblah · Answer

But why is that? As in how did you work it out

anonymous · Answer

oh, that's the multidimensional calculus answer to the differentiation assuming f(x,y) = 2x-y^2

hartnn · Answer

do you know chain rule ?

henryblah · Answer

Yes, but I don't see how it would work here

anonymous · Answer

when we differentiate with respect to 'x', then 'y' will be zero. and when we differentiate with repect to 'y', then 'x' will be consider as zero

hartnn · Answer

$\Large \dfrac{\partial u }{\partial x}$

means differentiating u w.r.t 'x' and treating 'y' as CONSTANT.

hartnn · Answer

$\Large \dfrac{\partial u }{\partial y}$
means differentiating u w.r.t 'y' and treating 'x' as CONSTANT.

henryblah · Answer

Yes, but how you differentiate something like f(2x) then? if you don't know what f is.

henryblah · Answer

oh hmm chain rule right?

hartnn · Answer

derivative of f(2x)  would be f'(2x) d/dx (2x)
simply put, chain rule.

hartnn · Answer

yes :)

henryblah · Answer

Ok, i'll try it and see if it works.

hartnn · Answer

$\Large [f(2x-y^4)]' = f'(2x-y^4) [2x-y^4]' $

henryblah · Answer

Hmm, that's the step that I think i'm getting wrong. I get du/dx to be
-e^-xf(2x-y^4) + 2e^-xf'(2x). But I think that f'(2x) should be f'(2x-y^4)

hartnn · Answer

thats correct.
-e^-xf(2x-y^4) + 2e^-xf'(2x-y^4)

henryblah · Answer

Oh right, I think it works!

henryblah · Answer

Thanks for the help.

hartnn · Answer

welcome ^_^