Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
3) z=y^2 cdot tg(x), x=t^2 cdot u cdot v, y=u+t cdot v^2 ; partial zpartial t, partial zpartial u, partial zpartial v quando t=2, u=1, v=0.

Passo 1

O enunciado dá
z = y^2,tan(x), x = t^2uv, y = u + t,v^2
e pede partial zpartial t, partial zpartial u e partial zpartial v em t=2, u=1, v=0. Vamos aplicar a Regra da Cadeia, usando
partial zpartial t = zx,xt + zy,yt, partial zpartial u = zx,xu + zy,yu, partial zpartial v = zx,xv + zy,yv

Passo 2

Derivadas parciais de z em relação a x e y:
zx = partialpartial xl(y^2tan(x)r) = y^2,sec^2(x),
zy = partialpartial yl(y^2tan(x)r) = 2y,tan(x)

Passo 3

Derivadas de x e y em relação a t,u,v:
x = t^2 u v Longrightarrow xt = 2tuv,;;xu = t^2v,;;xv = t^2u;
y = u + t,v^2 Longrightarrow yt = v^2,;;yu = 1,;;yv = 2t,v

Passo 4

Escrevendo as fórmulas gerais pela Regra da Cadeia:
partial zpartial t = zx,xt + zy,yt, partial zpartial u = zx,xu + zy,yu, partial zpartial v = zx,xv + zy,yv

Passo 5

Avaliar tudo em t=2, u=1, v=0:
x(2,1,0)=2^2cdot1cdot0=0, y(2,1,0)=1+2cdot0^2=1;
xt=2cdot2cdot1cdot0=0, xu=4cdot0=0, xv=4cdot1=4;
yt=0^2=0, yu=1, yv=2cdot2cdot0=0;
zx = 1^2sec^2(0)=1, zy = 2cdot1cdottan(0)=0
Substituindo:
partial zpartial t = 1cdot0 + 0cdot0 = 0, partial zpartial u = 1cdot0 + 0cdot1 = 0, partial zpartial v = 1cdot4 + 0cdot0 = 4

Resposta

partial zpartial t=0, partial zpartial u=0, partial zpartial v=4

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