Ache as derivadas parciais primeiras de f.
w=\sqrt{x^{2}+y^{2}+z^{2}}

Passo 1

O enunciado pede para encontrarmos as derivadas parciais primeiras de
w = \sqrt{x^{2}+y^{2}+z^{2}},
ou seja, \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y} e \frac{\partial w}{\partial z}. Vamos usar a regra da cadeia, considerando a função interna u(x,y,z)=x^{2}+y^{2}+z^{2} e a externa w(u)=u^{1/2}.

Passo 2

Escrevemos
w = (x^{2}+y^{2}+z^{2})^{\tfrac12}.
Pela regra da cadeia,
\frac{\partial w}{\partial x} = \frac{d}{dx}\bigl(u^{1/2}\bigr)\bigg|_{u=x^{2}+y^{2}+z^{2}} = \frac{1}{2}\,u^{-1/2}\,\frac{\partial u}{\partial x}.

Passo 3

Calculamos \frac{\partial u}{\partial x}:
\frac{\partial }{\partial x}(x^{2}+y^{2}+z^{2}) = 2x.
Substituindo em \frac{\partial w}{\partial x}:
\frac{\partial w}{\partial x} = \frac{1}{2}\,(x^{2}+y^{2}+z^{2})^{-1/2}\cdot 2x = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}.

Passo 4

Por simetria, fazendo o mesmo para y e z, obtemos
\frac{\partial w}{\partial y} = \frac{1}{2}\,(x^{2}+y^{2}+z^{2})^{-1/2}\cdot 2y = \frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},
\frac{\partial w}{\partial z} = \frac{1}{2}\,(x^{2}+y^{2}+z^{2})^{-1/2}\cdot 2z = \frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}.

Resposta

\frac{\partial w}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},\qquad \frac{\partial w}{\partial y} = \frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\qquad \frac{\partial w}{\partial z} = \frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}.

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