Calcule partial fpartial x(x, y) e partial fpartial y(x, y) para as funções abaixo:
f(x, y)=integralx+y^x^2-y^2 1+e^t d t

Passo 1

O enunciado nos dá a função
f(x,y)=integralx+y^,x^2-y^2l(1+e^tr),dt
e pede as derivadas parciais partial fpartial x e partial fpartial y.
O conteúdo aqui é a regra de Leibniz para derivada de integral com limites que dependem de x e y.

Passo 2

Definimos os limites
a(x,y)=x+y, b(x,y)=x^2-y^2
Calculamos as derivadas desses limites:
partial apartial x=1, partial bpartial x=2x, partial apartial y=1, partial bpartial y=-2y

Passo 3

Pela regra de Leibniz,
partial fpartial x =l(1+e^b(x,y)r)partial bpartial x -l(1+e^a(x,y)r)partial apartial x
Substituindo a, b e suas derivadas, obtemos
partial fpartial x =l(1+e^,x^2-y^2r)cdot(2x) -l(1+e^,x+yr)cdot1

Passo 4

De forma análoga, para partial fpartial y vale
partial fpartial y =l(1+e^b(x,y)r)partial bpartial y -l(1+e^a(x,y)r)partial apartial y,
logo
partial fpartial y =l(1+e^,x^2-y^2r)cdot(-2y) -l(1+e^,x+yr)cdot1

Resposta

partial fpartial x(x,y) =2xl(1+e^,x^2-y^2r)-l(1+e^,x+yr),
partial fpartial y(x,y) =-2yl(1+e^,x^2-y^2r)-l(1+e^,x+yr)

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