Questão 11. Calcular as derivadas parciais das seguintes funções:
h(x, y, z, w)=\frac{x+y z \cos \left(x w^{3}\right)}{x+z w}

Passo 1

O enunciado pede as derivadas parciais de
h(x,y,z,w)=\frac{x + y\,z\cos\bigl(xw^{3}\bigr)}{x + z\,w}\,.
O conteúdo envolvido é regra do quociente para derivadas de funções de várias variáveis, além de produto e cadeia ao derivar y z\cos(xw^3).

Passo 2

Definimos
N(x,y,z,w)=x + y\,z\cos\bigl(xw^{3}\bigr), \qquad D(x,z,w)=x + z\,w.
Para \displaystyle\frac{\partial h}{\partial x} usamos
\frac{\partial N}{\partial x} =1 + y\,z\bigl(-\sin(xw^3)\bigr)\cdot\frac{\partial(xw^3)}{\partial x} =1 - y\,z\,w^3\sin\bigl(xw^3\bigr),
\frac{\partial D}{\partial x}=1.
Logo
\frac{\partial h}{\partial x} =\frac{D\,\frac{\partial N}{\partial x}-N\,\frac{\partial D}{\partial x}}{D^2} =\frac{(x+zw)\bigl[1 - yzw^3\sin(xw^3)\bigr]\;-\;\bigl[x + yz\cos(xw^3)\bigr]}{(x+zw)^2}.

Passo 3

Para \displaystyle\frac{\partial h}{\partial y} notamos que D não depende de y e
\frac{\partial N}{\partial y} =z\cos\bigl(xw^3\bigr), \quad \frac{\partial D}{\partial y}=0.
Assim
\frac{\partial h}{\partial y} =\frac{D\,\frac{\partial N}{\partial y}}{D^2} =\frac{z\cos(xw^3)}{\,x+zw\,}.

Passo 4

Para \displaystyle\frac{\partial h}{\partial z} temos
\frac{\partial N}{\partial z} =y\cos\bigl(xw^3\bigr), \qquad \frac{\partial D}{\partial z}=w.
Aplicando a regra do quociente,
\frac{\partial h}{\partial z} =\frac{(x+zw)\,y\cos(xw^3)\;-\;w\bigl[x + yz\cos(xw^3)\bigr]}{(x+zw)^2}.

Passo 5

Para \displaystyle\frac{\partial h}{\partial w} calculamos
\frac{\partial N}{\partial w} =y\,z\bigl(-\sin(xw^3)\bigr)\,\frac{\partial(xw^3)}{\partial w} =-3\,x\,y\,z\,w^2\sin(xw^3),
\frac{\partial D}{\partial w}=z.
Logo
\frac{\partial h}{\partial w} =\frac{(x+zw)\bigl[-3\,x\,y\,z\,w^2\sin(xw^3)\bigr]\;-\;z\bigl[x + yz\cos(xw^3)\bigr]}{(x+zw)^2}.

Resposta

\displaystyle\frac{\partial h}{\partial x} =\frac{(x+zw)\bigl[1 - yzw^3\sin(xw^3)\bigr] - \bigl[x + yz\cos(xw^3)\bigr]}{(x+zw)^2}
\displaystyle\frac{\partial h}{\partial y} =\frac{z\cos(xw^3)}{x+zw}
\displaystyle\frac{\partial h}{\partial z} =\frac{(x+zw)\,y\cos(xw^3) - w\bigl[x + yz\cos(xw^3)\bigr]}{(x+zw)^2}
\displaystyle\frac{\partial h}{\partial w} =\frac{(x+zw)\bigl[-3\,x\,y\,z\,w^2\sin(xw^3)\bigr] - z\bigl[x + yz\cos(xw^3)\bigr]}{(x+zw)^2}

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