Math 22 Partial Derivatives

Partial Derivatives of a Function of Two Variables

 If $z=f(x,y)$, then the first partial derivatives of  with respect to $x$ and $y$ are the functions ${\frac {\partial z}{\partial x}}$ and ${\frac {\partial z}{\partial x}}$, defined as shown.
 
 ${\frac {\partial z}{\partial x}}=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x,y)-f(x,y)}{\Delta x}}$
 
 ${\frac {\partial z}{\partial y}}=\lim _{\Delta y\to 0}{\frac {f(x,y+\Delta y)-f(x,y)}{\Delta y}}$
 
 We can denote ${\frac {\partial z}{\partial x}}$ as $f_{x}(x,y)$ and ${\frac {\partial z}{\partial y}}$ as $f_{y}(x,y)$

Example: Find ${\frac {\partial z}{\partial x}}$ and ${\frac {\partial z}{\partial y}}$ of:

1) $z=f(x,y)=2x^{2}-4xy$

Solution:
${\frac {\partial z}{\partial x}}=4x^{2}-4y$
${\frac {\partial z}{\partial y}}=-4x$

2) $z=f(x,y)=x^{2}y^{3}$

Solution:
${\frac {\partial z}{\partial x}}=2xy^{3}$
${\frac {\partial z}{\partial y}}=3x^{2}y^{2}$

3) $z=f(x,y)=x^{2}e^{x^{2}y}$

Solution:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {\partial z}{\partial x}}=2xe^{x^{2}y}+x^{2}e^{x^{2}y}2xy} (product rule +chain rule)
${\frac {\partial z}{\partial y}}=x^{2}e^{x^{2}y}(x^{2})=x^{4}e^{x^{2}y}$

Higher-Order Partial Derivatives

1. ${\frac {\partial }{\partial x}}({\frac {\partial f}{\partial x}})={\frac {\partial ^{2}f}{\partial x^{2}}}=f_{xx}$ Return to Topics Page

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