Determine as derivadas parciais de primeira ordem das funções definidas por
w=\frac{x^{2} e^{y z}}{\tan \left(x z-y^{2}\right)}.

Passo 1

O enunciado define
w(x,y,z) = \frac{x^{2}e^{yz}}{\tan\bigl(xz - y^{2}\bigr)}
e pede as derivadas parciais de primeira ordem \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y} e \frac{\partial w}{\partial z}.
Para isso vamos usar a regra do quociente
\frac{\partial}{\partial v}\Bigl(\frac{N}{D}\Bigr) = \frac{N_v D - N\,D_v}{D^2}
e a regra da cadeia sempre que aparecer \tan(u) ou e^{yz}.

Passo 2

Definimos
N(x,y,z) = x^2 e^{yz}, \quad D(x,y,z) = \tan\bigl(u\bigr), \quad u = xz - y^2.
Então
N_x = 2x e^{yz}, \quad D_x = \operatorname{sec}^2(u)\,\frac{\partial u}{\partial x} = \operatorname{sec}^2(u)\,z.
Aplicando a regra do quociente,
\frac{\partial w}{\partial x} = \frac{2x e^{yz}\,\tan(u) \;-\; x^2 e^{yz}\,\operatorname{sec}^2(u)\,z} {\tan^2(u)}.

Passo 3

Para \partial w/\partial y, calculamos
N_y = x^2\,e^{yz}\,z, \quad D_y = \operatorname{sec}^2(u)\,\frac{\partial u}{\partial y} = \operatorname{sec}^2(u)\,(-2y).
Logo
\frac{\partial w}{\partial y} = \frac{x^2 z e^{yz}\,\tan(u) \;-\; x^2 e^{yz}\,(-2y)\operatorname{sec}^2(u)} {\tan^2(u)} = \frac{x^2 e^{yz}}{\tan^2(u)}\Bigl(z\,\tan(u) + 2y\,\operatorname{sec}^2(u)\Bigr).

Passo 4

Para \partial w/\partial z, temos
N_z = x^2\,e^{yz}\,y, \quad D_z = \operatorname{sec}^2(u)\,\frac{\partial u}{\partial z} = \operatorname{sec}^2(u)\,x.
Portanto
\frac{\partial w}{\partial z} = \frac{x^2 y e^{yz}\,\tan(u) \;-\; x^2 e^{yz}\,x\,\operatorname{sec}^2(u)} {\tan^2(u)} = \frac{x^2 e^{yz}}{\tan^2(u)}\Bigl(y\,\tan(u) - x\,\operatorname{sec}^2(u)\Bigr).

Resposta

\frac{\partial w}{\partial x} = \frac{2x e^{yz}\,\tan(xz - y^2)\;-\;x^2 e^{yz}\,z\,\operatorname{sec}^2(xz - y^2)} {\tan^2(xz - y^2)},
\frac{\partial w}{\partial y} = \frac{x^2 e^{yz}}{\tan^2(xz - y^2)}\Bigl(z\,\tan(xz - y^2) + 2y\,\operatorname{sec}^2(xz - y^2)\Bigr),
\frac{\partial w}{\partial z} = \frac{x^2 e^{yz}}{\tan^2(xz - y^2)}\Bigl(y\,\tan(xz - y^2) - x\,\operatorname{sec}^2(xz - y^2)\Bigr).

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