Encontre a derivada parcial indicada usando a regra da cadeia:
u=sen^-1(3 x+y) ; x=r^2 e^s ; y=sen(r s) ; partial upartial r ; partial upartial s

Passo 1

Resumo do enunciado: temos
u=sen^-1(3x+y),
x=r^2e^s,
y=sen(r,s),
e queremos
partial upartial r e partial upartial s.
Para isso vamos usar a regra da cadeia para derivadas parciais, além das regras de derivação de exponencial e de seno.

Passo 2

Derivamos primeiro a função externa sen^-1(z). Se z=3x+y, então
dudz=ddzl[sen^-1(z)r] =1raiz1-z^2 e partial zpartial r=3,partial xpartial r+partial ypartial r,
partial zpartial s=3,partial xpartial s+partial ypartial s

Passo 3

Calculamos as derivadas parciais de x e y:
x=r^2e^s ;Longrightarrow; partial xpartial r=2r,e^s, partial xpartial s=r^2e^s;
y=sen(r,s) ;Longrightarrow; partial ypartial r=scos(r,s), partial ypartial s=rcos(r,s)

Passo 4

Agora montamos partial upartial r pela regra da cadeia:
partial upartial r =1raiz1-(3x+y)^2 ;l(3,partial xpartial r+partial ypartial rr) =1raiz1-l(3r^2e^s+sen(r,s)r)^2 l(3cdot2r,e^s+scos(r,s)r)

Passo 5

Analogamente, para partial upartial s temos:
partial upartial s =1raiz1-(3x+y)^2 ;l(3,partial xpartial s+partial ypartial sr) =1raiz1-l(3r^2e^s+sen(r,s)r)^2 l(3,r^2e^s+rcos(r,s)r)

Resposta

partial upartial r =6r,e^s+scos(r,s) raiz1-l(3r^2e^s+sen(r,s)r)^2 , partial upartial s =3r^2e^s+rcos(r,s) raiz1-l(3r^2e^s+sen(r,s)r)^2

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