Para resolver a questão, precisamos calcular as derivadas parciais da função \( w = f(x, y, z) = x^2 y^3 z + 3x \) e depois avaliar no ponto \( P(1, 2, -1) \). 1. Derivadas parciais: - \( W_x = \frac{\partial w}{\partial x} = 2xy^3z + 3 \) - \( W_y = \frac{\partial w}{\partial y} = 3x^2y^2z \) - \( W_z = \frac{\partial w}{\partial z} = x^2y^3 \) 2. Derivadas de segunda ordem: - \( W_{xx} = \frac{\partial^2 w}{\partial x^2} = 2y^3z \) - \( W_{yy} = \frac{\partial^2 w}{\partial y^2} = 6x^2yz \) - \( W_{xz} = \frac{\partial^2 w}{\partial x \partial z} = 2xy^3 \) - \( W_{yz} = \frac{\partial^2 w}{\partial y \partial z} = 3x^2y^2 \) 3. Avaliar as derivadas no ponto \( P(1, 2, -1) \): - \( W_x(1, 2, -1) = 2(1)(2^3)(-1) + 3 = 2(1)(8)(-1) + 3 = -16 + 3 = -13 \) - \( W_{yy}(1, 2, -1) = 6(1^2)(2)(-1) = 6(1)(2)(-1) = -12 \) 4. Calcular \( W_x + W_{yy} + W_{xz} - W_{yz} \): - \( W_{xz}(1, 2, -1) = 2(1)(2^3) = 2(1)(8) = 16 \) - \( W_{yz}(1, 2, -1) = 3(1^2)(2^2) = 3(1)(4) = 12 \) 5. Resultado final: \[ W_x + W_{yy} + W_{xz} - W_{yz} = -13 - 12 + 16 - 12 = -21 \] Portanto, o valor obtido é -21.