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Definitions and Examples Partial derivatives help us track the change of multivariable
functions by dealing with one variable at a time. If we
think of $z=f(x,y)$ as being in 3space, we can discuss movement
in the $x$direction, or $y$direction, and see how this movement
affects $z$. We do this by holding $y$
fixed, and varying $x$, or vice versa.
Holding one variable fixed has the effect of slicing a cross section of 3space, which is
then 2space and we can use our knowledge to
understand it. Definitions and notation

The definition of
partial derivatives The partial derivative of $f$ with respect to $x$ is $\displaystyle{ \lim_{h \rightarrow 0} \frac{f(x+h,y)f(x,y)}{h}}.$ The partial derivative of $f$ with respect to $y$ is $\displaystyle{ \lim_{h \rightarrow 0} \frac{f(x,y+h)f(x,y)}{h}}.$ In general we do not use these definitions to compute partial derivatives. 
There are many notations for partial derivatives. If $z =
f(x,y)$, then some, but not all, of the notations:
The partial derivative of $f$ with respect to $x$:
$\displaystyle f_x(x,y) = f_x = \frac{\partial f}{\partial x}=
\frac{\partial}{\partial x}f(x,y) = \frac{\partial z}{\partial x}=
f_1$.
The partial derivative of $f$ with respect to $y$: $\displaystyle f_y(x,y) = f_y = \frac{\partial f}{\partial y}= \frac{\partial}{\partial y}f(x,y) = \frac{\partial z}{\partial y}= f_2$.
We can evaluate these partial derivatives at particular values of
$x$ and $y$, e.g. $f_x(2,7)=\frac{\partial f}{\partial
x}\big_{(2,7)}$ or $f_y(1,10)$.
To find $f_x$, hold $y$ constant and differentiate with respect
to $x$. To find $f_y$, hold $x$ constant and differentiate
with respect to $y$. Literally, when computing $f_y$ we
treat $x$ as a constant because it is
a constant. This appears as a slice of 3space. We can
slice 3 dimensions at a particular $x$value, say we slice at $x=1$ (see the
previous page for a graphic example); such a slice is parallel to
the $yz$plane. The $x$value is the same everywhere in this slice
 it's constant. Then we observe what happens to $z$ as $y$
changes. This procedure makes computing partial derivatives
very simple.

Example 2: Compute $f_x$ and $f_y$ when $f(x,y) =
\sin(x+y^2)$.
Solution 2: We must use the chain rule here.
Since the derivative with respect to $x$ of $\sin(x + \hbox{
constant })$ is $\cos(x + \hbox{ constant })\cdot(1+0)$, and since
the derivative with respect to $y$ of $\sin(\hbox{constant} +
y^2)$ is $ \cos(\hbox{constant }+y^2)\cdot(0+2y)$, we get
$\displaystyle f_x = \cos(x+y^2)$ and $\displaystyle f_y = 2y
\cos(x+y^2).$

Example 3: Compute both partial derivatives
of $\tan{\left(xy^2+7\right)}$
Example 4: Find $\displaystyle\frac{\partial
f}{\partial x},\frac{\partial f}{\partial y}$ for
$f(x,y)=\sqrt{x^25y}\left(\ln{xy}\right)$.
Example 5: Find $f_x(1,2)$ and $f_y(1,2)$ for
$f(x,y)=3x^24y^37x^2y^3$ from Example 1 above.
What do these numbers mean?
DO: Try to compute these
derivatives before looking ahead.
Solution 3: We must use the chain rule.
$\displaystyle\frac{\partial}{\partial
x}(\tan(xy^2+7))=\sec^2(xy^2+7)\frac{\partial}{\partial
x}(xy^2+7)=\sec^2(xy^2+7)(y^2)$.
$\displaystyle\frac{\partial}{\partial
y}(\tan(xy^2+7))=\sec^2(xy^2+7)\frac{\partial}{\partial
y}(xy^2+7)=\sec^2(xy^2+7)(2xy)$.

Solution 4: We must use the product rule and the
chain rule:
$\displaystyle\frac{\partial f}{\partial
x}=\frac12(x^25y)^{1/2}(2x0)\ln(xy)+\sqrt{x^25y}\
\frac{y}{xy}=\frac{x}{\sqrt{x^25y}}\ln(xy)+\frac{\sqrt{x^25y}}{x}$.
$\displaystyle\frac{\partial f}{\partial
y}=\frac12(x^25y)^{1/2}(05)\ln(xy)+\sqrt{x^25y}\
\frac{x}{xy}=\frac{5\ln(xy)}{2\sqrt{x^25y}}+\frac{\sqrt{x^25y}}{y}$.

Solution 5: Since $f_x(x,y)=6x14xy^3$,
$f_x(1,2)=6(1)14(1)2^3=6+112=106$. And since
$f_y(x,y)=12y^221x^2y^2$, $f_y(1,2)=12\cdot
2^221(1)^2\cdot 2^2=4884=48$. This
means that if we stand at the point $(1,2)$ and
look in the positive $x$ direction, $z=f(x,y)$ is heading upward,
but if we look in the positive $y$ direction $z$ is heading
downward.