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When to Use the Quotient Rule for Differentiation

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  1. 0:06 Introduction to the Quotient Rule
  2. 1:18 Understanding the Quotient Rule
  3. 4:09 Quotient Rule Examples
  4. 7:29 Lesson Summary
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Taught by

Erin Lennon

Erin has taught math and science from grade school up to the post-graduate level. She holds a Ph.D. in Chemical Engineering.

Lo D Hi minus Hi D Lo, all over the square of what's below! Learn the quotient rule chant for differentiating functions that take the form of fractions in this lesson.

Introduction to the Quotient Rule

The quotient rule is the last of the main rules for calculating derivatives, and it primarily deals with what happens if you have a function divided by another function and you want to take the derivative of that. So let's start with f(x) = x / x^2. What is the derivative of f(x)? Is it just the top? Is it the bottom? Is it the derivative of the top times the derivative of the bottom? What is it? Well you could simplify this into 1 / x, and then f`(x) would then be -1 / x^2, because 1 / x is the same as x^-1. Then we just use our power rule. But is there a way to find that without simplifying? You won't be able to simplify in every case. Think about sin(x) / x, or (x^2 + ln(x)) / cos(x). What about those cases? You need to know the quotient rule.

Using the quotient rule in example #1
Example 1 Quotient Rule

Understanding the Quotient Rule

Let's say that you have y=u / v, where both u and v depend on x. Then you want to find dy/dx, or d/dx(u / v). There are two ways to find that. One is to use the power rule, then the product rule, then the chain rule. First you redefine u / v as uv^-1. Then you're going to differentiate; y` is the derivative of uv^-1. You need to use the product rule. So you've got y`= u(d/dx)v^-1 + v^-1(d/dx)u. Then we need to use the chain rule to differentiate v^-1, so y`= u(-1(1 / v^2)v`) + (v^-1)u`. I can rewrite this as y`= (u` / v) - (uv` / v^2). I can put all of this together by multiplying the left term by (v / v), so y` ends up being (vu` - uv`) / v^2. And that's what happens if you try to use the power rule, the product rule and the chain rule to differentiate u/v.

The second way to differentiate u / v is to use the quotient rule, which has this nice little jingle: Low d hi minus hi d low, all over the square of what's below. Here, u is the high and v is the low, so y`= (vu` - uv`) / v^2, and you end up with the same equation for y`. So if you can remember this jingle, this is definitely the way to go.

Differentiating using the quotient rule in example #2
Quotient Rule Example 2

Quotient Rule Examples

Let's use this for our example: x / x^2. x is my high, and x^2 is my low. So (x^2(d/dx)x - x(d/dx)x^2) / (x^2)^2. Well, the derivative of x with respect to x is just 1, and the derivative of x^2 with respect to x is 2x. If I plug all of this in and simplify, I get -x^2 / x^4, or -1/x^2, which is exactly what I had before.

Let's use this for a slightly more complex case: y = 3x / (x + x^2). y` = ((x + x^2)(3) - (3x)(1 + 2x)) / (x + x^2)^2. I can simplify this by expanding out these two terms and canceling these 3xs.

That wasn't so bad, but what about something like f(x) = ((x + x^3)^2) / sin(4x)? Here I'm going to do something a little bit different and write out exactly what the lows and highs are.

The final example problem
Quotient Rule Example 3
Low is sin(4x), so 'd' low is 4cos(4x). High is (x + x^3)^2, and 'd' high (which is a little more complex because we have to use the chain rule) = 2(x + x^3)(1 + 3x^2). Now that I've got that written out, I can just go through our jingle. f`(x) = ((sin(4x) * 2(x + x^3)(1 + 3x^2)) - ((x + x^3)^2 * 4cos(4x)) / (sin(4x))^2.

Lesson Summary

Let's review. You want to use the quotient rule when you have one function divided by another function and you're taking the derivative of that, such as u / v. And you can remember the quotient rule by remembering this little jingle: Lo d hi minus hi d low, all over the square of what's below.

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