Use a Regra da Cadeia para determinar fracpartial zpartial t e fracpartial zpartial s.
z=e^x y tan (y), x=s+2 t, y=st

Passo 1

O enunciado define
z = e^x y,tan(y), x = s + 2t, y = st,
e pede partial z/partial s e partial z/partial t. O conteúdo aqui é a Regra da Cadeia para derivadas parciais de funções compostas.

Passo 2

Primeiro, calculamos as derivadas parciais de z em relação a x e a y.
Como
z = e^xy,tan(y),
temos
partial zpartial x = partialpartial x(e^xy);tan(y) = e^xy,y,tan(y),
partial zpartial y = partialpartial y(e^xy);tan(y) + e^xy,ddy(tan(y)) = e^xy,x,tan(y) + e^xy,sec^2(y) = e^xyl(x,tan(y)+sec^2(y)r)

Passo 3

Agora as derivadas de x e y em relação a s e a t:
x = s + 2t Longrightarrow partial xpartial s=1,;partial xpartial t=2,
y = st Longrightarrow partial ypartial s=1t,;partial ypartial t=-st^2

Passo 4

pela Regra da Cadeia:
partial zpartial s = partial zpartial x,partial xpartial s + partial zpartial y,partial ypartial s,
partial zpartial t = partial zpartial x,partial xpartial t + partial zpartial y,partial ypartial t

Passo 5

Substituindo:
partial zpartial s = e^xyytan(y)cdot1 + e^xyl(xtan(y)+sec^2(y)r)cdot1t = e^xy!l(ytan(y)+xtan(y)+sec^2(y)tr),
partial zpartial t = e^xyytan(y)cdot2 + e^xyl(xtan(y)+sec^2(y)r)cdotl(-st^2r) = e^xy!l(2ytan(y)-s,(xtan(y)+sec^2(y))t^2r)

Resposta

partial zpartial s = e^xy!l(ytan(y)+xtan(y)+sec^2(y)tr),
partial zpartial t = e^xy!l(2ytan(y)-s,(xtan(y)+sec^2(y))t^2r)

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