Se f(x, y)=(x-y) \sin (3 x+2 y). Calcule f_{x}\left(0, \frac{\pi}{3}\right) e f_{y}\left(0, \frac{\pi}{3}\right).

Passo 1

O enunciado dá a função
f(x,y) = (x - y)\,\sin\bigl(3x + 2y\bigr)
e pede as derivadas parciais \displaystyle f_{x} e \displaystyle f_{y} no ponto \bigl(0,\tfrac{\pi}{3}\bigr). Vamos usar regra do produto e regra da cadeia para calcular as derivadas parciais.

Passo 2

Para obter \displaystyle f_{x}(x,y), aplicamos produto entre u(x,y)=x-y e v(x,y)=\sin(3x+2y):
f_{x} = \frac{\partial u}{\partial x}\,v + u\,\frac{\partial v}{\partial x}
Aqui \partial u/\partial x = 1 e, pela cadeia,
\frac{\partial v}{\partial x} = \cos\bigl(3x+2y\bigr)\,\frac{\partial}{\partial x}(3x+2y) = 3\,\cos\bigl(3x+2y\bigr).
Logo,
f_{x}(x,y) = \sin\bigl(3x+2y\bigr) + (x - y)\,3\,\cos\bigl(3x+2y\bigr).

Passo 3

Agora avaliamos em \bigl(0,\tfrac{\pi}{3}\bigr). Note que
3\cdot 0 + 2\cdot \tfrac{\pi}{3} = \tfrac{2\pi}{3}, \quad \sin\!\bigl(\tfrac{2\pi}{3}\bigr) = \tfrac{\sqrt{3}}{2}, \quad \cos\!\bigl(\tfrac{2\pi}{3}\bigr) = -\tfrac12, \quad x - y = 0 - \tfrac{\pi}{3} = -\tfrac{\pi}{3}.
Substituindo,
f_{x}\Bigl(0,\tfrac{\pi}{3}\Bigr) = \tfrac{\sqrt{3}}{2} + \bigl(-\tfrac{\pi}{3}\bigr)\,3\,\bigl(-\tfrac12\bigr) = \tfrac{\sqrt{3}}{2} + \tfrac{\pi}{2} = \frac{\sqrt{3} + \pi}{2}.

Passo 4

Para \displaystyle f_{y}(x,y), usamos de novo o produto:
f_{y} = \frac{\partial u}{\partial y}\,v + u\,\frac{\partial v}{\partial y}.
Temos \partial u/\partial y = -1 e, pela cadeia,
\frac{\partial v}{\partial y} = \cos\bigl(3x+2y\bigr)\,\frac{\partial}{\partial y}(3x+2y) = 2\,\cos\bigl(3x+2y\bigr).
Portanto
f_{y}(x,y) = -\sin\bigl(3x+2y\bigr) + (x - y)\,2\,\cos\bigl(3x+2y\bigr).

Passo 5

No ponto \bigl(0,\tfrac{\pi}{3}\bigr), usando os mesmos valores de seno e cosseno,
f_{y}\Bigl(0,\tfrac{\pi}{3}\Bigr) = -\tfrac{\sqrt{3}}{2} + 2\bigl(-\tfrac{\pi}{3}\bigr)\bigl(-\tfrac12\bigr) = -\tfrac{\sqrt{3}}{2} + \tfrac{\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}.

Resposta

f_{x}\Bigl(0,\tfrac{\pi}{3}\Bigr) = \frac{\sqrt{3} + \pi}{2}, \qquad f_{y}\Bigl(0,\tfrac{\pi}{3}\Bigr) = \frac{\pi}{3} - \frac{\sqrt{3}}{2}.

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