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 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.

Definitions and notation

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 $f_y(1,-10)$.

Computing partial derivatives

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.

--------------------------------------------------------------------------------

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 $f(x,y)=\sqrt{x^2-5y}\left(\ln{xy}\right)$.

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.
                        $\displaystyle\frac{\partial}{\partial x}(\tan(xy^2+7))=\sec^2(xy^2+7)\frac{\partial}{\partial x}(xy^2+7)=\sec^2(xy^2+7)(y^2)$.

                        $\displaystyle\frac{\partial}{\partial y}(\tan(xy^2+7))=\sec^2(xy^2+7)\frac{\partial}{\partial y}(xy^2+7)=\sec^2(xy^2+7)(2xy)$.

--------------------------------------------------------------------------------

Solution 4:  We must use the product rule and the chain rule:
                $\displaystyle\frac{\partial f}{\partial x}=\frac12(x^2-5y)^{-1/2}(2x-0)\ln(xy)+\sqrt{x^2-5y}\
\frac{y}{xy}=\frac{x}{\sqrt{x^2-5y}}\ln(xy)+\frac{\sqrt{x^2-5y}}{x}$.

                $\displaystyle\frac{\partial f}{\partial y}=\frac12(x^2-5y)^{-1/2}(0-5)\ln(xy)+\sqrt{x^2-5y}\
\frac{x}{xy}=-\frac{5\ln(xy)}{2\sqrt{x^2-5y}}+\frac{\sqrt{x^2-5y}}{y}$.

--------------------------------------------------------------------------------

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.