Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
T=\frac{v}{2 u+v}, \quad u=p q \sqrt{r}, \quad v=p \sqrt{q} r; \frac{\partial T}{\partial p}, \frac{\partial T}{\partial q}, \frac{\partial T}{\partial r} quando p=2, q=1, r=4

Passo 1

O enunciado define
T=\frac{v}{2u+v},\quad u=pq\sqrt{r},\quad v=p\sqrt{q}\,r
e pede \frac{\partial T}{\partial p}, \frac{\partial T}{\partial q}, \frac{\partial T}{\partial r} em p=2, q=1, r=4. Vamos usar a Regra da Cadeia, que diz por exemplo
\frac{\partial T}{\partial p} =\frac{\partial T}{\partial u}\,\frac{\partial u}{\partial p} +\frac{\partial T}{\partial v}\,\frac{\partial v}{\partial p}.

Passo 2

Calcular as derivadas parciais de T em relação a u e v. Como
T(v,u)=\frac{v}{2u+v}=v\,(2u+v)^{-1},
temos
\frac{\partial T}{\partial u} =v\cdot(-1)\,(2u+v)^{-2}\cdot2 =-\frac{2v}{(2u+v)^2},
\frac{\partial T}{\partial v} =(2u+v)^{-1}+v\cdot(-1)\,(2u+v)^{-2}\cdot1 =\frac{2u}{(2u+v)^2}.

Passo 3

Calcular as derivadas parciais de u e v em relação a p,q,r:
u(p,q,r)=p\,q\,\sqrt{r} \quad\implies\quad \frac{\partial u}{\partial p}=q\sqrt{r},\; \frac{\partial u}{\partial q}=p\sqrt{r},\; \frac{\partial u}{\partial r}=p\,q\cdot\frac{1}{2\sqrt{r}}.
v(p,q,r)=p\,\sqrt{q}\,r \quad\implies\quad \frac{\partial v}{\partial p}=\sqrt{q}\,r,\; \frac{\partial v}{\partial q}=p\,r\cdot\frac{1}{2\sqrt{q}},\; \frac{\partial v}{\partial r}=p\sqrt{q}.

Passo 4

Avaliar todas as quantidades em p=2, q=1, r=4. Primeiro
u=2\cdot1\cdot2=4,\quad v=2\cdot1\cdot4=8, \quad2u+v=8+8=16, \quad(2u+v)^2=256.
Logo
\frac{\partial T}{\partial u} =-\frac{2\cdot8}{256}=-\frac{16}{256}=-\frac{1}{16}, \quad \frac{\partial T}{\partial v} =\frac{2\cdot4}{256}=\frac{8}{256}=\frac{1}{32}.
E
\frac{\partial u}{\partial p}=1\cdot2=2,\; \frac{\partial u}{\partial q}=2\cdot2=4,\; \frac{\partial u}{\partial r}=\frac{2\cdot1}{2\cdot2}=\frac12,
\frac{\partial v}{\partial p}=1\cdot4=4,\; \frac{\partial v}{\partial q}=\frac{2\cdot4}{2\cdot1}=4,\; \frac{\partial v}{\partial r}=2\cdot1=2.

Passo 5

Aplicar a Regra da Cadeia:
\displaystyle\frac{\partial T}{\partial p} =\frac{\partial T}{\partial u}\,\frac{\partial u}{\partial p} +\frac{\partial T}{\partial v}\,\frac{\partial v}{\partial p} =-\tfrac{1}{16}\cdot2+\tfrac{1}{32}\cdot4 =0.
\displaystyle\frac{\partial T}{\partial q} =-\tfrac{1}{16}\cdot4+\tfrac{1}{32}\cdot4 =-\tfrac{1}{4}+\tfrac{1}{8} =-\tfrac{1}{8}.
\displaystyle\frac{\partial T}{\partial r} =-\tfrac{1}{16}\cdot\tfrac12+\tfrac{1}{32}\cdot2 =-\tfrac{1}{32}+\tfrac{2}{32} =\tfrac{1}{32}.

Resposta

\displaystyle\frac{\partial T}{\partial p}=0,\quad \frac{\partial T}{\partial q}=-\tfrac{1}{8},\quad \frac{\partial T}{\partial r}=\tfrac{1}{32}.

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