Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
w=(x+y+z)^2, x=r-s, y=cos (r+s), y=sen(r+s);
partial wpartial r, partial wpartial s, quando r=1, s=-1

Passo 1

O enunciado define
w = (x+y+z)^2,
com
x = r - s, y = cos(r+s), z = sin(r+s)
Queremos usar a Regra da Cadeia para calcular partial w/partial r e partial w/partial s e depois avaliar em r=1,;s=-1.

Passo 2

Primeiro, obtemos as derivadas parciais de x,y,z em relação a r e a s:
partial xpartial r = 1, partial xpartial s = -1
partial ypartial r = -sin(r+s), partial ypartial s = -sin(r+s)
partial zpartial r = cos(r+s), partial zpartial s = cos(r+s)

Passo 3

Defina u = x+y+z, de modo que w = u^2. Pela Regra da Cadeia:
partial wpartial r = 2,u,partial upartial r, partial wpartial s = 2,u,partial upartial s,
onde
partial upartial r = partial xpartial r + partial ypartial r + partial zpartial r, partial upartial s = partial xpartial s + partial ypartial s + partial zpartial s

Passo 4

Avalie em r=1,;s=-1. Então r+s=0, sin0=0, cos0=1, e
u(1,-1) = (1-(-1)) + cos0 + sin0 = 2 + 1 + 0 = 3
Em seguida,
partial upartial r(1,-1) = 1 + l(-sin0r) + cos0 = 1 + 0 + 1 = 2,
partial upartial s(1,-1) = -1 + l(-sin0r) + cos0 = -1 + 0 + 1 = 0
Logo
partial wpartial r(1,-1) = 2cdot3cdot2 = 12,
partial wpartial s(1,-1) = 2cdot3cdot0 = 0

Resposta

partial wpartial r(1,-1)=12, partial wpartial s(1,-1)=0.

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