Proof of Power Rule for 2 variables

ak123456
Messages
50
Reaction score
0

Homework Statement



u^n (x,y)=nu^(n-1) (x,y) u' (x,y)

Homework Equations





The Attempt at a Solution


can i set f (x,y)=u^n (x,y)
lnf =lnu^n
lnf=nlnU
f'/f=n/U
f'=fn/U -(U(^n) )(n/U)
after that i don't know how to continue and is there a better way to prove it
 
Physics news on Phys.org
ak123456 said:

Homework Statement



u^n (x,y)=nu^(n-1) (x,y) u' (x,y)
What do you mean by "u'(x,y)"? The gradient? The differential? "u' " is not defined for a function of two variables. If you mean the gradient, which is what I would think of as "the" derivative for a function of several variables, then yes, \nabla u^n(x,y)= nu^{n-1}\nabla u. That follows from the chain rule.

Homework Equations





The Attempt at a Solution


can i set f (x,y)=u^n (x,y)
lnf =lnu^n
lnf=nlnU
f'/f=n/U
f'=fn/U -(U(^n) )(n/U)
after that i don't know how to continue and is there a better way to prove it
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top