Proof of Power Rule for 2 variables

  • Thread starter ak123456
  • Start date
1. The problem statement, all variables and given/known data

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

2. Relevant equations



3. 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 dont know how to continue and is there a better way to prove it
 

HallsofIvy

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1. The problem statement, all variables and given/known data

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, [itex]\nabla u^n(x,y)= nu^{n-1}\nabla u[/itex]. That follows from the chain rule.

2. Relevant equations



3. 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 dont know how to continue and is there a better way to prove it
 

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