# Difference between revisions of "Math 22 Partial Derivatives"

## Partial Derivatives of a Function of Two Variables

 If ${\displaystyle z=f(x,y)}$, then the first partial derivatives of  with respect to ${\displaystyle x}$ and ${\displaystyle y}$ are the functions ${\displaystyle {\frac {\partial z}{\partial x}}}$ and ${\displaystyle {\frac {\partial z}{\partial x}}}$, defined as shown.

${\displaystyle {\frac {\partial z}{\partial x}}=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x,y)-f(x,y)}{\Delta x}}}$

${\displaystyle {\frac {\partial z}{\partial y}}=\lim _{\Delta y\to 0}{\frac {f(x,y+\Delta y)-f(x,y)}{\Delta y}}}$

We can denote ${\displaystyle {\frac {\partial z}{\partial x}}}$ as ${\displaystyle f_{x}(x,y)}$ and ${\displaystyle {\frac {\partial z}{\partial y}}}$ as ${\displaystyle f_{y}(x,y)}$


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

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

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

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

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

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

Solution:
${\displaystyle {\frac {\partial z}{\partial x}}=2xe^{x^{2}y}+x^{2}e^{x^{2}y}2xy}$ (product rule +chain rule)
${\displaystyle {\frac {\partial z}{\partial y}}=x^{2}e^{x^{2}y}(x^{2})=x^{4}e^{x^{2}y}}$

## Higher-Order Partial Derivatives

1. ${\displaystyle {\frac {\partial }{\partial x}}({\frac {\partial f}{\partial x}})={\frac {\partial ^{2}f}{\partial x^{2}}}=f_{xx}}$

2. ${\displaystyle {\frac {\partial }{\partial y}}({\frac {\partial f}{\partial y}})={\frac {\partial ^{2}f}{\partial y^{2}}}=f_{yy}}$

3. ${\displaystyle {\frac {\partial }{\partial y}}({\frac {\partial f}{\partial x}})={\frac {\partial ^{2}f}{\partial y\partial x}}=f_{xy}}$

4. ${\displaystyle {\frac {\partial }{\partial x}}({\frac {\partial f}{\partial y}})={\frac {\partial ^{2}f}{\partial x\partial y}}=f_{yx}}$

1) Find ${\displaystyle f_{xy}}$, given that ${\displaystyle f(x,y)=2x^{2}-4xy}$,

Solution:
${\displaystyle f_{x}=4x-4y}$
Then, ${\displaystyle f_{xy}=-4}$

2) Find ${\displaystyle f_{yx}}$, given that ${\displaystyle z=f(x,y)=3xy^{2}-2y+5x^{2}y^{2}}$,

Solution:
${\displaystyle f_{y}=6xy-2+10x^{2}y}$
Then, ${\displaystyle f_{yx}=6y+20xy}$