Encontre a derivada parcial indicada usando a regra da cadeia:
u=cosh (yx) ; x=2 s e^r ; y=6 s e^r ; partial upartial r ; partial upartial s

Passo 1

No enunciado, temos a função
u = cosh!l(yxr)
com
x = 2,s,e^r, y = 6,s,e^r,
e queremos calcular as derivadas parciais partial u/partial r e partial u/partial s. O conteúdo principal é a regra da cadeia e, dentro dela, a regra do quociente para tfracyx.

Passo 2

Pela regra da cadeia,
partial upartial r =sinh!l(tfracyxr),partialpartial rl(tfracyxr), partial upartial s =sinh!l(tfracyxr),partialpartial sl(tfracyxr)

Passo 3

Calculamos primeiro partial x/partial r e partial y/partial r:
partial xpartial r = 2,s,e^r = x, partial ypartial r = 6,s,e^r = y
Pela regra do quociente,
partialpartial rl(tfracyxr) =x,partial ypartial r;-;y,partial xpartial rx^2 =x,y - y,xx^2 =0

Passo 4

Como partial(y/x)/partial r = 0, segue
partial upartial r =sinh!l(tfracyxr)cdot 0 =0

Passo 5

Agora partial x/partial s e partial y/partial s:
partial xpartial s=2,e^r, partial ypartial s=6,e^r
De novo pela regra do quociente,
partialpartial sl(tfracyxr) =x,(6e^r) - y,(2e^r)x^2 =2s e^rcdot6e^r - 6s e^rcdot2e^rx^2 =0

Passo 6

Como partial(y/x)/partial s = 0, concluímos
partial upartial s =sinh!l(tfracyxr)cdot 0 =0

Resposta

partial upartial r = 0, partial upartial s = 0

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