#### Video introduction

*Partial derivative examples.*

Once you understand the concept of a partial derivative as the rate that something is changing, calculating partial derivatives usually isn't difficult. (Unfortunately, there are special cases where calculating the partial derivatives is hard.) As these examples show, calculating a partial derivatives is usually just like calculating an ordinary derivative of one-variable calculus. You just have to remember with which variable you are taking the derivative.

#### Example 1

Let $f(x,y) = y^3x^2$. Calculate $\displaystyle \pdiff{f}{x}(x,y)$.

**Solution**: To calculate $\displaystyle \pdiff{f}{x}(x,y)$, we simply view
$y$ as being a fixed number and calculate the ordinary derivative with
respect to $x$. The first time you do this, it might be easiest to
set $y=b$, where $b$ is a constant, to remind you that you should
treat $y$ as though it were number rather than a variable. Then, the
partial derivative $\displaystyle \pdiff{f}{x}(x,y)$ is the same as
the ordinary derivative of the function $g(x)=b^3x^2$. Using the rules
for ordinary differentiation, we know that
\begin{align*}
\diff{g}{x}(x) = 2b^3x.
\end{align*}
Now, we remember that $b=y$ and substitute $y$ back in to conclude that
\begin{align*}
\pdiff{f}{x}(x,y) = 2y^3x.
\end{align*}

#### Example 2

For the same $f$, calculate $\displaystyle \pdiff{f}{y}(x,y)$.

**Solution**: This time, we'll just calculate the derivative with respect
to $y$ directly without replacing $x$ with a constant. We just have
to remember to treat $x$ like a constant and use the rules for
ordinary differentiation. We don't touch the $x^2$ and only
differentiate the $y^3$ factor to calculate that
\begin{align*}
\pdiff{f}{y}(x,y) =3 x^2 y^2.
\end{align*}

#### Example 3

For the same $f$, calculate $\displaystyle \pdiff{f}{x}(1,2)$.

**Solution**: From example 1, we know that $\displaystyle
\pdiff{f}{x}(x,y) = 2y^3x$. To evaluate this partial derivative at
the point $(x,y)=(1,2)$, we just substitute the respective values for
$x$ and $y$:
\begin{align*}
\pdiff{f}{x}(1,2) = 2 (2^3)(1) = 16.
\end{align*}

#### Example 4

For \begin{align*} f(x_1,x_2,x_3,x_4) = 3 \frac{\cos(x_1x_4)\sin(x_2^5)}{e^{x_2}+ (1+x_2^2)/(x_1x_2x_4)} + 5x_1x_3x_4 \end{align*} calculate $\displaystyle \pdiff{f}{x_3}(a,b,c,d)$.

**Solution**: Although this initially looks hard, it's really any easy
problem. The ugly term does not depend on $x_3$, so in calculating
partial derivative with respect to $x_3$, we treat it as a constant.
The derivative of a constant is zero, so that term drops out. The
derivative is just the derivative of the last term with respect to
$x_3$, which is
\begin{align*}
\pdiff{f}{x_3}(x_1,x_2,x_3,x_4) = 5x_1x_4
\end{align*}
Substituting in the values $(x_1,x_2,x_3,x_4)=(a,b,c,d)$, we obtain
the final answer
\begin{align*}
\pdiff{f}{x_3}(a,b,c,d) = 5ad.
\end{align*}

#### Example 5

Let \begin{align*} p(y_1,y_2,y_3) = 9\frac{y_1y_2y_3}{y_1+y_2+y_3} \end{align*} and calculate $\displaystyle \pdiff{p}{y_3}(y_1,y_2,y_3)$ at the point $(y_1,y_2,y_3)=(1,-2,4)$.

**Solution**: In calculating partial derivatives, we can use all the rules for ordinary derivatives. We can calculate $\pdiff{p}{y_3}$ using the quotient rule.
\begin{align*}
\pdiff{p}{y_3}(y_1,y_2,y_3) &= 9\frac{\displaystyle(y_1+y_2+y_3)\pdiff{}{y_3}(y_1y_2y_3)
-(y_1y_2y_3)\pdiff{}{y_3}(y_1+y_2+y_3) }{(y_1+y_2+y_3)^2}\\
&= 9\frac{(y_1+y_2+y_3)(y_1y_2)-(y_1y_2y_3)1 }{(y_1+y_2+y_3)^2}\\
&= 9\frac{(y_1+y_2)y_1y_2}{(y_1+y_2+y_3)^2}.
\end{align*}

Plugging in the point $(y_1,y_2,y_3)=(1,-2,4)$ yields the answer \begin{align*} \pdiff{p}{y_3}(1,-2,4) &= 9\frac{(1-2)1(-2)}{(1-2+4)^2}= 2. \end{align*}