The Derivative, Part 6

Last time I presented the multiplication rule of differentiation to be used when given a function that is the multiplication of two or more other functions. As you would guess, there is also a rule that handles the division of two functions.

Let’s say you have the function

\[ f( x) =\frac{x^{2}}{\text{sin}( x)}\]

This one can be solved with the multiplication rule if you remember that 1/sin(x) = csc(x). But as I haven’t told you what the derivative of csc(x) is, we are stuck using the following division rule. But this highlights the point that as we get deeper into maths, there are often several ways to solve a problem. The maths “arteest” is one that solves a problem elegantly.

So the following rule is the division rule. Again, I will use u(x) and v(x) to split the function up into its parts. If you have a function of the form

\[f( x) =\frac{u( x)}{v( x)}\]

then the derivative of f(x) is

\[f'( x) =\frac{u( x) v'( x) -u'( x) v( x)}{[ v( x)]^{2}}\]

As you can see, this rule is a bit more complex which is why you would use a simpler rule if possible. But it is still relatively easy to use if you keep track of which part is u and which part is v.

Using the example function above,

\[\begin{array}{{>{\displaystyle}l}} u( x) =x^{2} ,\ \ \ \ \ v( x) =\text{sin}( x)\\ u'( x) =2x,\ \ \ \ v'( x) =\text{cos}( x) \end{array}\]

So according to the division rule,

\[f'( x) =\frac{x^{2}\text{cos}( x) -2x\text{sin}( x)}{\text{sin}^{2}( x)}\]

Now you can use many rules in a single differentiation problem consider

\[f( x) =\frac{x^{2} e^{x}}{\text{sin}( x)}\]

Here, the numerator is a multiplication of two functions. So when using the division rule, you need to apply the multiplication rule for the u‘ part:

\[ \begin{array}{{>{\displaystyle}l}}
u( x) \ =\ x^{2} e^{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v( x) =\text{sin}( x)\\
u'( x) =x^{2} e^{x} +2xe^{x} \ \ \ \ \ v'( x) =\text{cos}( x)

I’ll leave it as an exercise for you to see if I correctly found u‘(x) using the multiplication rule. I used the fact that (as seen from the table I provided a couple of posts before) is its own derivative. Anyway, using the division rule,

\[f'( x) =\frac{x^{2} e^{x}\text{cos}( x) -\left( x^{2} e^{x} +2xe^{x}\right)\text{sin}( x)}{\text{sin}^{2}( x)}\]

So you might be thinking that you can differentiate any function as long as you know the derivatives of the individual parts. So how would you differentiate

\[ f( x) =\text{sin}\left( x^{2}\right) ?\]

This is not a multiplication of functions, but rather a function of a function. I will introduce the very powerful chain rule as it applies to differentiation in my next post.