The Quotient Rule

Similar to the product rule, the quotient rule is a tool for finding complex derivatives by breaking them down into simpler pieces.  There is one main difference between the two.  As the names imply, the product rule is applicable when you need to find the derivative of a function that is actually the product of two simpler functions and the quotient rule is used when your function can be described as one simpler function being divided by another.

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Like the product rule, there are a few different ways you might see the quotient rule represented.  I recommend picking the one that makes the most sense to you so that you can memorize the formula.

For our functions f(x) and g(x):

$$\frac{d}{dx}\Bigg[\frac{f(x)}{g(x)}\Bigg]=\frac{\frac{d}{dx}\big[f(x)\big]g(x)-f(x)\frac{d}{dx}\big[g(x)\big]}{\big(g(x)\big)^{2}}$$

$$\Bigg(\frac{f}{g}\Bigg)’=\frac{f’\cdot g-f\cdot g’}{g^2}$$

$$\Bigg(\frac{f(x)}{g(x)}\Bigg)’=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$$

Unlike the product rule, the order does matter here.  This is because we are using subtraction and division rather than addition and multiplication.  Since we need the pieces to be in the correct order, it is helpful to come up with some method for memorizing the quotient rule as I have shown it above.

How to remember the quotient rule

We need to take the derivative of some function, that can be represented as a fraction made up of two functions that are easier to derive.  In other words we will be considering the “top” of the fraction and the “bottom” of the fraction as two separate, simpler functions.  The method I like to use to remember this formula is to think of the function in the numerator and the “high” portion and the denominator as the “low” portion.  These will be abbreviated as “Hi” and “Lo.”

$$\Bigg(\frac{f(x)}{g(x)}\Bigg)’=\Bigg(\frac{Hi}{Lo}\Bigg)’=\frac{Lo \ dHi-Hi \ dLo}{Lo^2}$$

And if you read it out loud, it almost seems to have a little jingle to it, which makes is easier to remember:

Lo d Hi minus Hi d Lo, all over Lo squared.

Just to add a little clarification, the “d” means you should take the derivative of the following piece.  So, “d Hi” should be where you plug in the derivative of the function that makes up the top half of the fraction.

Example 1

Find the derivative of h(x)=\frac{e^x}{\sqrt{x}}.

Recognizing when to use the quotient rule

Just like we did with the product rule example, we want to first recognize how we will split up this function.  In other words, we need to recognize which part we will consider f(x) and which part we will consider g(x).  The main difference is that this distinction does matter with the quotient rule.  We need to call f(x) our top function and g(x) our bottom function.

Therefore, we need to say f(x)=e^x and g(x)=\sqrt{x}.  Once we have made this distinction, we can consider these two functions individually for a moment and find each of their derivatives.  I already found these derivatives in Example 1 of The Product Rule.  If you want to see this again, click here.

By using the previous example, we already know f'(x) and g'(x).  I recommend writing out f(x), f'(x), g(x), and g'(x) all in one place before using the quotient rule formula.  So far we have:

$$f(x)=e^x$$

$$g(x)=\sqrt{x}=x^\frac{1}{2}$$

$$f'(x)=e^x$$

$$g'(x)=\frac{1}{2}x^{-\frac{1}{2}}$$

Putting it all together

Now that we have figured out these four parts, we can simply plug them into the quotient rule formula we have above.  This tells us that:

$$h'(x)=\Bigg(\frac{e^x}{\sqrt{x}}\Bigg)’=\frac{\Big(x^\frac{1}{2}\Big)(e^x)-(e^x)\Big(\frac{1}{2}x^{-\frac{1}{2}}\Big)}{\big(\sqrt{x}\big)^2}=\frac{e^x\Big(x^\frac{1}{2}-\frac{1}{2}x^{-\frac{1}{2}}\Big)}{x}$$

$$=e^x\Big(x^{-\frac{1}{2}}-\frac{1}{2}x^{-\frac{3}{2}}\Big)$$

Work with me

Please don’t forget to leave a comment or email me at jakesmathlessons@gmail.com with any questions or suggestions for future lessons!  Also, I recommend going to the derivatives page to see other topics I’ve covered.  And you can see more quotient rule practice problems here.  Just send me an email if you can’t find what you’re looking for and I’ll see what I can do to help!

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