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 $\displaystyle \pdiff(x,y)$.

Solution. To calculate $\displaystyle \pdiff(x,y)$, 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 $\displaystyle \pdiff(x,y)$ 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 \diff(x) = 2b^3x. \end Now, we remember that $b=y$ and substitute $y$ back in to conclude that \begin \pdiff(x,y) = 2y^3x. \end

Example 2

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

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 \pdiff(x,y) =3 x^2 y^2. \end

Example 3

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

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

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 \pdiff(x_1,x_2,x_3,x_4) = 5x_1x_4 \end Substituting in the values $(x_1,x_2,x_3,x_4)=(a,b,c,d)$, we obtain the final answer \begin \pdiff(a,b,c,d) = 5ad. \end

Example 5

Let \begin p(y_1,y_2,y_3) = 9\frac \end and calculate $\displaystyle \pdiff(y_1,y_2,y_3)$ 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 $\pdiff$ using the quotient rule. \begin \pdiff(y_1,y_2,y_3) &= 9\frac<\displaystyle(y_1+y_2+y_3)\pdiff<>(y_1y_2y_3) -(y_1y_2y_3)\pdiff<>(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 Plugging in the point $(y_1,y_2,y_3)=(1,-2,4)$ yields the answer \begin \pdiff(1,-2,4) &= 9\frac<(1-2)1(-2)><(1-2+4)^2>= 2. \end