Derivadas parciales de orden superior para tres variables. Cálculo vectorial.

Problemas resueltos.

Problema 1. Hallar las derivadas parciales de orden superior para la siguiente función:

$\displaystyle f(x,y,z) = ye^x + x \ln{z}$

Solución. Derivando parcialmente con respecto a x

$\displaystyle {f}_{x} (x, y,z) = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (ye^x + x \ln{z})$

$\displaystyle {f}_{x} (x,y,z) = \frac{\partial}{\partial x} (ye^x ) + \frac{\partial}{\partial x} (x \ln{z}) = y \frac{\partial}{\partial x} (e^x) + \ln{z} \frac{\partial}{\partial x} (x)$

$\displaystyle \therefore {f}_{x} (x,y,z) = ye^x + \ln{z}$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{y} (x,y,z) = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (y e^x + x \ln{z})$

$\displaystyle {f}_{y} (x,y,z) = \frac{\partial}{\partial y} (ye^x) + \frac{\partial}{\partial y} (x \ln{z}) = e^x \frac{\partial}{\partial y} (y) + x \ln{z} \frac{\partial}{\partial y} (1)$

$\therefore {f}_{y} (x,y) = e^x$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{z} (x,y,z) = \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (y e^x + x \ln{z})$

$\displaystyle {f}_{y} (x,y,z) = \frac{\partial}{\partial z} (ye^x) + \frac{\partial}{\partial z} (x \ln{z}) = y{e}^{x} \frac{\partial}{\partial z} (1) + x \frac{\partial}{\partial z} (\ln{z})$

$\displaystyle \therefore {f}_{z} (x,y,z) = \frac{x}{z}$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{xx} (x,y,z) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x}) = \frac{{\partial}^{2} f}{\partial x^2} = \frac{\partial}{\partial x} (ye^x + \ln{z})$

$\displaystyle {f}_{xx} (x,y,z) = \frac{\partial}{\partial x} (ye^x) + \frac{\partial}{\partial x} (\ln{z}) = y \frac{\partial}{\partial x} (e^x) + \ln{z} \frac{\partial}{\partial x} (1)$

$\displaystyle \therefore {f}_{xx} (x,y,z) = ye^x$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{xy} (x,y,z) = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) = \frac{{\partial}^{2} f}{\partial y \partial x} = \frac{\partial}{\partial y} (ye^x + \ln{z})$

$\displaystyle {f}_{xy} (x,y,z) = \frac{\partial}{\partial y} (ye^x) + \frac{\partial}{\partial y} (\ln{z}) = {e}^{x} \frac{\partial}{\partial y} (y) + \ln{z} \frac{\partial}{\partial y} (1)$

$\therefore {f}_{xy} (x,y,z) = e^x$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{xz} (x,y,z) = \frac{\partial}{\partial z} (\frac{\partial f}{\partial x}) = \frac{{\partial}^{2} f}{\partial z \partial x} = \frac{\partial}{\partial z} (ye^x + \ln{z})$

$\displaystyle {f}_{xz} (x,y,z) = \frac{\partial}{\partial z} (ye^x) + \frac{\partial}{\partial z} (\ln{z}) = ye^x \frac{\partial}{\partial z} (1) + \frac{\partial}{\partial z} (\ln{z})$

$\displaystyle \therefore {f}_{xz} (x,y,z) = \frac{1}{z}$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{yx} (x,y,z) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial y}) = \frac{{\partial}^{2} f}{\partial x \partial y} = \frac{\partial}{\partial x} (e^x)$

$\therefore {f}_{yx} (x,y,z) = e^x$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{yy} (x,y,z) = \frac{\partial}{\partial y} (\frac{\partial f}{\partial y}) = \frac{{\partial}^{2} f}{\partial y^2} = \frac{\partial}{\partial y} (e^x) = e^x \frac{\partial}{\partial y} (1)$

$\therefore {f}_{yy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{yz} (x,y,z) = \frac{\partial}{\partial z} (\frac{\partial f}{\partial y}) = \frac{{\partial}^{2} f}{\partial z \partial y} = \frac{\partial}{\partial z} (e^x) = e^x \frac{\partial}{\partial z} (1)$

$\therefore {f}_{yz} (x,y,z) = 0$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{zx} (x,y,z) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial z}) = \frac{{\partial}^{2} f}{\partial x \partial z} = \frac{\partial}{\partial x} (\frac{x}{z}) = \frac{1}{z} \frac{\partial}{\partial x} (x)$

$\displaystyle \therefore {f}_{zx} (x,y,z) = \frac{1}{z}$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{zy} (x,y,z) = \frac{\partial}{\partial y} (\frac{\partial f}{\partial z}) = \frac{{\partial}^{2} f}{\partial y \partial z} = \frac{\partial}{\partial y} (\frac{x}{z}) = \frac{x}{z} \frac{\partial}{\partial y} (1)$

$\therefore {f}_{zy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{zz} (x,y,z) = \frac{\partial}{\partial z} (\frac{\partial f}{\partial z}) = \frac{{\partial}^{2} f}{\partial z^2} = x \frac{\partial}{\partial z} (\frac{1}{z}) = x \frac{\partial}{\partial z} ({z}^{-1}) = x(-{z}^{-2})$

$\displaystyle \therefore {f}_{zz} (x,y,z) = -\frac{x}{z^2}$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{xxx} (x,y,z) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x^2}) = \frac{{\partial}^{3} f}{\partial x^3} = \frac{\partial}{\partial x} (ye^x) = y \frac{\partial}{\partial x} (e^x)$

$\therefore {f}_{xxx} (x,y,z) = ye^x$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{xxy} (x,y,z) = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x^2}) = \frac{{\partial}^3 f}{\partial y \partial x^2} = \frac{\partial}{\partial y} (ye^x) = e^x \frac{\partial}{\partial x} (y)$

$\therefore {f}_{xxy} (x,y,z) = e^x$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{xxz} (x,y,z) = \frac{\partial}{\partial z} (\frac{\partial f}{\partial x^2}) = \frac{{\partial}^{3} f}{\partial z \partial x^2} = \frac{\partial}{\partial z} (ye^x) = ye^x \frac{\partial}{\partial x} (1)$

$\therefore {f}_{xxz} (x,y,z) = 0$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{xyx} (x,y,z) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial y \partial x}) = \frac{{\partial}^{3} f}{\partial y \partial x} = \frac{\partial}{\partial x} (e^x)$

$\therefore {f}_{xyx} (x,y,z) = e^x$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{xyy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial y \partial x}) = \frac{{\partial}^{3} f}{\partial y^2 \partial x} = \frac{\partial}{\partial y} (e^x) = e^x \frac{\partial}{\partial y} (1)$

$\therefore {f}_{xyy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{xyz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{\partial y \partial x}) = \frac{{\partial}^{3} f}{\partial z \partial y \partial x} = \frac{\partial}{\partial z} (e^x) = e^x \frac{\partial}{\partial z} (1)$

$\therefore {f}_{xyz}(x,y,z) = 0$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{xzx} (x,y,z) = \frac{\partial}{\partial x} (\frac{\partial f}{\partial z \partial x}) = \frac{{\partial}^{3} f}{\partial x^2 \partial z} = \frac{partial}{\partial x} (\frac{1}{z}) = \frac{1}{z} \frac{\partial}{\partial x} (1)$

$\therefore {f}_{xzx} (x,y,z) = 0$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{xzy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial z \partial x}) = \frac{{\partial}^{3} f}{\partial y \partial z \partial x} = \frac{\partial}{\partial y} (\frac{1}{z}) = \frac{1}{z} \frac{\partial}{\partial y} (1)$

$\therefore {f}_{xzy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{xzz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{\partial x \partial x}) = \frac{{\partial}^{3} f}{\partial z^2 \partial x} = \frac{\partial}{\partial z} (\frac{1}{z}) = \frac{\partial}{\partial z} ({z}^{-1}) = -{z}^{-2} = -\frac{1}{z^2}$

$\displaystyle \therefore {f}_{xzz} (x,y,z) = -\frac{1}{z^2}$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{yxx} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial x \partial y}) = \frac{{\partial}^{3} f}{\partial x \partial y^2} = \frac{\partial}{\partial x} (e^x)$

$\therefore {f}_{yxx} (x,y,z) = e^x$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{yxy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial x \partial y}) = \frac{{\partial}^{3} f}{\partial y^2 \partial x} = \frac{\partial}{\partial y} (e^x) = e^x \frac{\partial}{\partial y} (1)$

$\therefore {f}_{yxy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{yxz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{\partial y \partial x}) = \frac{{\partial}^{3} f}{\partial z \partial y \partial x} = \frac{\partial}{\partial z} (e^x) = e^x \frac{\partial}{\partial z} (1)$

$\therefore {f}_{yxz} (x,y,z) = 0$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{yyx} (x,y,z) = \frac{\partial}{\partial x} (\frac{{\partial}^{2} f}{\partial y^2}) = \frac{{\partial}^{3} f}{\partial x \partial y^2} = \frac{\partial}{\partial x} (0)$

$\therefore {f}_{yyx} (x,y,z) = 0$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{yyy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^2 f}{{\partial} y^2}) = \frac{{\partial}^{3} f}{\partial y^3} = \frac{\partial}{\partial y} (0)$

$\therefore {f}_{yyy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{yyz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{{\partial}y^{2}}) = \frac{{\partial}^{3} f}{\partial z \partial y^2} = \frac{\partial}{\partial z} (0)$

$\therefore {f}_{yyz} (x,y,z) = 0$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{yzx} (x,y,z) = \frac{\partial}{\partial x} (\frac{{\partial}^{2} f}{\partial z \partial y}) = \frac{{\partial}^{3} f}{\partial x \partial z \partial y} = \frac{\partial}{\partial x} (0)$

$\therefore {f}_{yzx} (x,y,z) = 0$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{yzy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial z \partial y}) = \frac{{\partial}^{3} f}{\partial z \partial y^2} = \frac{\partial}{\partial y} (0)$

$\therefore {f}_{yzy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{yzz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{\partial z \partial y}) = \frac{{\partial}^{3} f}{\partial z^2 \partial y} = \frac{\partial}{\partial z} (0)$

$\therefore {f}_{yzz} (x,y,z) = 0$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{zxx} (x,y,z) = \frac{\partial}{\partial x} (\frac{{\partial}^{2} f}{\partial x \partial z}) = \frac{{\partial}^{3} f}{\partial x^2 \partial z} = \frac{\partial}{\partial x} (\frac{1}{z}) = \frac{1}{z} \frac{\partial}{\partial x} (1)$

$\therefore {f}_{zxx} (x,y,z) = 0$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{zxy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial x \partial z}) = \frac{{\partial}^{3} f}{\partial y \partial x \partial z} = \frac{\partial}{\partial y} (\frac{1}{z}) = \frac{1}{z} \frac{\partial}{\partial y} (1)$

$\therefore {f}_{zxy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{zxz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{\partial x \partial z}) = \frac{{\partial}^{3} f}{\partial z^2 \partial x} = \frac{\partial}{\partial z} (\frac{1}{z}) = \frac{\partial}{\partial z} ({z}^{-1}) = -{z}^{-2} = -\frac{1}{z^2}$

$\displaystyle \therefore {f}_{zxz} (x,y,z) = -\frac{1}{z^2}$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{zyx} (x,y,z) = \frac{\partial}{\partial x} (\frac{{\partial}^{2} f}{\partial y \partial z}) = \frac{{\partial}^{3} f}{\partial x \partial y \partial z} = \frac{\partial}{\partial x} (0)$

$\therefore {f}_{zyx} (x,y,z) = 0$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{zyy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial y \partial z}) = \frac{{\partial}^{3} f}{\partial y^2 \partial z} = \frac{\partial}{\partial y} (0)$

$\therefore {f}_{zyy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{zyz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{\partial y \partial z}) = \frac{{\partial}^{3} f}{\partial z^2 \partial y} = \frac{\partial}{\partial z} (0)$

$\therefore {f}_{zyz} (x,y,z) = 0$

Derivando parcialmente con respecto a x

$\displaystyle {f}_{zzx} (x,y,z) = \frac{\partial}{\partial x} (\frac{{\partial}^{2} f}{\partial z^2}) = \frac{{\partial}^{3} f}{\partial x \partial z^3} = \frac{\partial}{\partial x} (-\frac{x}{z^2}) = -\frac{1}{z^2} \frac{\partial}{\partial x} (x)$

$\therefore {f}_{zzx} (x,y,z) = -\frac{1}{z^2}$

Derivando parcialmente con respecto a y

$\displaystyle {f}_{zzy} (x,y,z) = \frac{\partial}{\partial y} (\frac{{\partial}^{2} f}{\partial z^2}) = \frac{{\partial}^{3} f}{\partial y \partial z^2} = \frac{\partial}{\partial y} (-\frac{x}{z^2}) = -\frac{x}{z^2} \frac{\partial}{\partial y} (1)$

$\therefore {f}_{zzy} (x,y,z) = 0$

Derivando parcialmente con respecto a z

$\displaystyle {f}_{zzz} (x,y,z) = \frac{\partial}{\partial z} (\frac{{\partial}^{2} f}{\partial z^2})$

$\displaystyle = \frac{{\partial}^{3} f}{\partial z^3} = \frac{\partial}{\partial z} (-\frac{x}{z^2}) = -x \frac{\partial}{\partial z} (\frac{1}{z^2}) = -x \frac{\partial}{\partial z} ({z}^{-2}) = -x(-2{z}^{-3}) = \frac{2x}{z^3}$

$\displaystyle \therefore {f}_{zzz} (x,y,z) = \frac{2x}{z^3}$

4-7DERIVADAS DE ORDEN SUPERIOR PARA TRES VARIABLES — *Figura 4.7.1 Imagen ilustrativa del resultado final de cada derivada parcial.*

Referencias bibliográficas.

Colley, S. J. (2013). Cálculo vectorial. México: PEARSON EDUCACIÓN.
Larson, R., & Edwards, B. (2017). Matemáticas 3. Cálculo de varias variables. México: CENGAGE Learning.
R. Spiegel, M. (1967). Análisis vectorial. México: McGRAW – HILL.

Derivadas parciales de orden superior para tres variables. Cálculo vectorial.

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Derivadas parciales de orden superior para tres variables. Cálculo vectorial.

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