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 3-space, 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 3-space, which is
then 2-space and we can use our knowledge to
understand it.
| 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 3-space. 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.
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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 $\displaystyle f_y = 2y
\cos(x+y^2).$
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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
$f(x,y)=\sqrt{x^2-5y}\left(\ln{xy}\right)$.
Example 5: Find $f_x(-1,2)$ and $f_y(-1,2)$ for
$f(x,y)=3x^2-4y^3-7x^2y^3$ 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.
$\displaystyle\frac{\partial}{\partial
x}(\tan(xy^2+7))=\sec^2(xy^2+7)\frac{\partial}{\partial
x}(xy^2+7)=\sec^2(xy^2+7)(y^2)$.
$\displaystyle\frac{\partial}{\partial
y}(\tan(xy^2+7))=\sec^2(xy^2+7)\frac{\partial}{\partial
y}(xy^2+7)=\sec^2(xy^2+7)(2xy)$.
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Solution 4: We must use the product rule and the
chain rule:
$\displaystyle\frac{\partial f}{\partial
x}=\frac12(x^2-5y)^{-1/2}(2x-0)\ln(xy)+\sqrt{x^2-5y}\
\frac{y}{xy}=\frac{x}{\sqrt{x^2-5y}}\ln(xy)+\frac{\sqrt{x^2-5y}}{x}$.
$\displaystyle\frac{\partial f}{\partial
y}=\frac12(x^2-5y)^{-1/2}(0-5)\ln(xy)+\sqrt{x^2-5y}\
\frac{x}{xy}=-\frac{5\ln(xy)}{2\sqrt{x^2-5y}}+\frac{\sqrt{x^2-5y}}{y}$.
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Solution 5: Since $f_x(x,y)=6x-14xy^3$,
$f_x(-1,2)=6(-1)-14(-1)2^3=-6+112=106$. And since
$f_y(x,y)=12y^2-21x^2y^2$, $f_y(-1,2)=12\cdot
2^2-21(-1)^2\cdot 2^2=48-84=-48$. 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.