Home»Math Guides»Partial Derivatives (multivariable quotient rule) – Example 3
Partial Derivatives Example 3 (Partial derviative of a fraction, use quotient rule)
In multivariable calculus you may be asked to find the partial derivatives.
When deriving with respect to a variable, just treat all other variables as a constant.
Let’s spice things up by finding partial derivatives of a fraction, which means you have to do partial derivatives using the quotient rule.
Find all first partial derivatives of \(f(x,y) = \frac{x}{{{{(x + y)}^2}}}\)
Now you need to use the quotient rule, but when doing this derive with respect to the variable you’re finding the partial derivative for!
Remember, the quotient rule is:
\[{\left[ {\frac{f}{g}} \right]^\prime } = \frac{{gf’ – g’f}}{{{g^2}}}\]
Where f is the numerator and g is the denominator of the function you need to find the derivative for.
But when finding f’ and g’, you do it with respect to the variable you’re finding the partial derivative for.
Let’s find the partial derivative with respect to x first, and we’ll show all examples since the quotient rule involves a bit of arithmetic.
Try to factor when possible to reduce the power of the denominator.
\[\frac{{\partial f}}{{\partial x}} = \frac{{{{(x + y)}^2}(1) – 2(x + y)x}}{{{{(x + y)}^4}}}\]
\[\frac{{\partial f}}{{\partial x}} = \frac{{(x + y – 2x)(x + y)}}{{{{(x + y)}^4}}}\]
\[\frac{{\partial f}}{{\partial x}} = \frac{{y – x}}{{{{(x + y)}^3}}}\]
There we go, it’s a bit of arithmetic but we should always try to simplify as much as reasonably possible.
Now, let’s find the partial derivative with respect to y, and the quotient rule is the same except our derivatives are found with respect to the “y” variable instead.
\[\frac{{\partial f}}{{\partial y}} = \frac{{{{(x + y)}^2}(0) – 2(x + y)x}}{{{{(x + y)}^4}}}\]
\[\frac{{\partial f}}{{\partial y}} = \frac{{ – 2x(x + y)}}{{{{(x + y)}^4}}}\]
\[\frac{{\partial f}}{{\partial y}} = \frac{{ – 2x}}{{{{(x + y)}^3}}}\]