The discussion centers on the proof of the Power Rule for functions of two variables, specifically the equation u^n(x,y) = nu^(n-1)(x,y)u'(x,y). Participants clarify that "u'" refers to the gradient of u, denoted as ∇u, and confirm that the derivative of u^n with respect to its variables follows from the chain rule. The correct application of the chain rule leads to the conclusion that ∇u^n(x,y) = nu^(n-1)∇u, establishing the validity of the Power Rule in multivariable calculus.
PREREQUISITESStudents and educators in multivariable calculus, mathematicians focusing on calculus proofs, and anyone seeking to understand the application of the Power Rule in functions of two variables.
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.ak123456 said: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