Find \nabla f at the given point.
f(x, y, z)=2 z^{3}-3\left(x^{2}+y^{2}\right) z+\tan ^{-1} x z, \quad(1,1,1)

Passo 1

O enunciado pede o gradiente de
f(x,y,z)=2z^{3}-3\bigl(x^{2}+y^{2}\bigr)z+\bigl(\tan^{-1}x\bigr)\,z
no ponto (1,1,1).
Conteúdo: derivadas parciais de uma função de três variáveis.

Passo 2

Calculamos \displaystyle \frac{\partial f}{\partial x}, tratando y,z como constantes.
  1. \partial_x[2z^3]=0.
  2. \partial_x[-3(x^2+y^2)z]=-3\cdot2x\cdot z=-6xz.
  3. \partial_x\bigl[(\tan^{-1}x)\,z\bigr]=z\cdot\frac{1}{1+x^2}.
    Portanto,
\frac{\partial f}{\partial x}=-6xz+\frac{z}{1+x^2}.

Passo 3

Calculamos \displaystyle \frac{\partial f}{\partial y}, tratando x,z como constantes.
  1. \partial_y[2z^3]=0.
  2. \partial_y[-3(x^2+y^2)z]=-3\cdot2y\cdot z=-6yz.
  3. \partial_y\bigl[(\tan^{-1}x)\,z\bigr]=0.
    Logo,
\frac{\partial f}{\partial y}=-6yz.

Passo 4

Calculamos \displaystyle \frac{\partial f}{\partial z}, tratando x,y como constantes.
  1. \partial_z[2z^3]=6z^2.
  2. \partial_z[-3(x^2+y^2)z]=-3(x^2+y^2).
  3. \partial_z\bigl[(\tan^{-1}x)\,z\bigr]=\tan^{-1}x.
    Portanto,
\frac{\partial f}{\partial z} =6z^2-3\bigl(x^2+y^2\bigr)+\tan^{-1}x.

Passo 5

Avaliaremos cada parcial em (x,y,z)=(1,1,1).
1.
\frac{\partial f}{\partial x}\Big|_{(1,1,1)} =-6\cdot1\cdot1+\frac{1}{1+1}=-6+\frac12=-\frac{11}{2}.
\frac{\partial f}{\partial y}\Big|_{(1,1,1)} =-6\cdot1\cdot1=-6.
\frac{\partial f}{\partial z}\Big|_{(1,1,1)} =6\cdot1^2-3(1^2+1^2)+\tan^{-1}1 =6-6+\frac{\pi}{4}=\frac{\pi}{4}.

Resposta

\nabla f(1,1,1) =\biggl(-\tfrac{11}{2},\,-6,\,\tfrac{\pi}{4}\biggr).

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