Dada u=sen(xy)+ln (yx). Mostre que y partial upartial y+x partial upartial x=0.

Passo 1

O enunciado define
u = sin!l(xyr) + ln!l(yxr)
e pede para mostrar que
y,partial upartial y ;+; x,partial upartial x ;=; 0
Para isso vamos usar cálculo de derivadas parciais, a regra da cadeia e as derivadas de sin e de ln.

Passo 2

Calcular partial upartial x.
Da soma temos
partialpartial xsin!l(tfracxyr) = cos!l(tfracxyr)cdotpartialpartial xl(tfracxyr) = 1y,cos!l(tfracxyr)
e
partialpartial xln!l(tfracyxr) = partialpartial x(ln y - ln x) = 0 - 1x = -1x
Logo,
partial upartial x = 1y,cos!l(tfracxyr) ;-; 1x

Passo 3

Calcular partial upartial y.
Para o sin vem
partialpartial ysin!l(tfracxyr) = cos!l(tfracxyr)cdotpartialpartial yl(tfracxyr) = cos!l(tfracxyr)cdotl(-tfracxy^2r) = -xy^2,cos!l(tfracxyr),
e para o logaritmo
partialpartial yln!l(tfracyxr) = partialpartial y(ln y - ln x) = 1y - 0 = 1y
Logo,
partial upartial y = -xy^2,cos!l(tfracxyr) ;+; 1y

Passo 4

Agora formamos
x,partial upartial x = xl(1ycos!tfracxy-1xr) = xycos!l(tfracxyr)-1,
y,partial upartial y = yl(-xy^2cos!tfracxy+1yr) = -xycos!l(tfracxyr)+1
Somando:
y,uy + x,ux = l(-tfracxycos!tfracxy+1r) +l(tfracxycos!tfracxy-1r) = 0

Resposta

0

Exercícios relacionados