Questão ( 20 pontos) Se \operatorname{tg}^{-1}(x z)+\operatorname{sen}(3 y+2 z)=\sqrt{x+y}, calcule \frac{\partial z}{\partial x} e \frac{\partial z}{\partial y}.

Passo 1

O enunciado dá a relação implícita
\arctan(x\,z)+\sin\bigl(3y+2z\bigr)=\sqrt{x+y}
e pede \partial z/\partial x e \partial z/\partial y.
O conteúdo é diferenciação implícita de função z=z(x,y). Iremos definir
F(x,y,z)=\arctan(xz)+\sin(3y+2z)-\sqrt{x+y}
e usar
F_x + F_z\,z_x=0, \quad F_y + F_z\,z_y=0.

Passo 2

Definimos
F(x,y,z)=\arctan(xz)+\sin(3y+2z)-\sqrt{x+y}.
Pelas fórmulas da diferenciação implícita, temos
\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}, \quad \frac{\partial z}{\partial y}=-\frac{F_y}{F_z}.

Passo 3

Agora calculamos as derivadas parciais de F:
F_x = \frac{\partial}{\partial x}\!\bigl[\arctan(xz)\bigr] - \frac{\partial}{\partial x}\!\bigl[\sqrt{x+y}\bigr] = \frac{z}{1+(xz)^2} - \frac{1}{2\sqrt{x+y}},
F_y = \frac{\partial}{\partial y}\!\bigl[\sin(3y+2z)\bigr] - \frac{\partial}{\partial y}\!\bigl[\sqrt{x+y}\bigr] = 3\cos(3y+2z) - \frac{1}{2\sqrt{x+y}},
F_z = \frac{\partial}{\partial z}\!\bigl[\arctan(xz)\bigr] + \frac{\partial}{\partial z}\!\bigl[\sin(3y+2z)\bigr] = \frac{x}{1+(xz)^2} + 2\cos(3y+2z).

Passo 4

Substituindo em
z_x=-\frac{F_x}{F_z}, \quad z_y=-\frac{F_y}{F_z},
obtemos:
\frac{\partial z}{\partial x} =-\frac{\displaystyle \frac{z}{1+(xz)^2}-\frac{1}{2\sqrt{x+y}}} {\displaystyle \frac{x}{1+(xz)^2}+2\cos(3y+2z)},
\frac{\partial z}{\partial y} =-\frac{\displaystyle 3\cos(3y+2z)-\frac{1}{2\sqrt{x+y}}} {\displaystyle \frac{x}{1+(xz)^2}+2\cos(3y+2z)}.

Resposta

\frac{\partial z}{\partial x} =-\frac{\dfrac{z}{1+(xz)^2}-\dfrac{1}{2\sqrt{x+y}}} {\dfrac{x}{1+(xz)^2}+2\cos(3y+2z)}, \quad \frac{\partial z}{\partial y} =-\frac{\;3\cos(3y+2z)-\dfrac{1}{2\sqrt{x+y}}\;} {\dfrac{x}{1+(xz)^2}+2\cos(3y+2z)}.

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