Determine as derivadas parciais indicadas.
f(x, y, z)=\sqrt{\cos ^{2} x+\cos ^{2} y+\cos ^{2} z} ; \quad f_{z}(\pi / 2, \pi / 2, \pi / 4)

Passo 1

O enunciado dá a função
f(x,y,z)=\sqrt{\cos^2 x + \cos^2 y + \cos^2 z}
e pede a derivada parcial em relação a z no ponto \bigl(\tfrac\pi2,\tfrac\pi2,\tfrac\pi4\bigr), ou seja, f_z(\tfrac\pi2,\tfrac\pi2,\tfrac\pi4). Vamos usar a regra da cadeia e a derivada de raiz quadrada.

Passo 2

Defina
u(x,y,z)=\cos^2 x + \cos^2 y + \cos^2 z,
então f=\sqrt{u}. Pela regra da cadeia,
f_z = \frac{1}{2\sqrt{u}}\;\frac{\partial u}{\partial z}.
Como \partial(\cos^2 z)/\partial z = 2\cos z\cdot(-\sin z) e \cos^2 x e \cos^2 y não dependem de z, temos
\frac{\partial u}{\partial z} = -2\cos z\,\sin z
e portanto
f_z(x,y,z) = \frac{-2\cos z\,\sin z}{2\sqrt{\cos^2 x + \cos^2 y + \cos^2 z}} = \frac{-\cos z\,\sin z}{\sqrt{\cos^2 x + \cos^2 y + \cos^2 z}}.

Passo 3

Agora avaliamos as funções trigonométricas no ponto \bigl(\tfrac\pi2,\tfrac\pi2,\tfrac\pi4\bigr):
\cos\bigl(\tfrac\pi2\bigr)=0,\quad \cos\bigl(\tfrac\pi4\bigr)=\frac{\sqrt2}{2},\quad \sin\bigl(\tfrac\pi4\bigr)=\frac{\sqrt2}{2}.
O denominador fica
\sqrt{\cos^2(\tfrac\pi2)+\cos^2(\tfrac\pi2)+\cos^2(\tfrac\pi4)} = \sqrt{0 + 0 + \Bigl(\tfrac{\sqrt2}{2}\Bigr)^2} = \sqrt{\tfrac12} = \frac{\sqrt2}{2}.

Passo 4

Substituindo em \displaystyle f_z=\frac{-\cos z\,\sin z}{\sqrt{\cos^2 x + \cos^2 y + \cos^2 z}},
obtemos
f_z\bigl(\tfrac\pi2,\tfrac\pi2,\tfrac\pi4\bigr) = \frac{-\bigl(\tfrac{\sqrt2}{2}\bigr)\bigl(\tfrac{\sqrt2}{2}\bigr)}{\tfrac{\sqrt2}{2}} = \frac{-\tfrac12}{\tfrac{\sqrt2}{2}} = -\frac{1}{\sqrt2} = -\frac{\sqrt2}{2}.

Resposta

f_z\Bigl(\tfrac\pi2,\tfrac\pi2,\tfrac\pi4\Bigr) = -\frac{\sqrt2}{2}.

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