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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Eric Garneau*

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.

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.

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.

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 2*x*. 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* = 3*x* / (*x* + *x*^2). *y`* = ((*x* + *x*^2)(3) - (3*x*)(1 + 2*x*)) / (*x* + *x*^2)^2. I can simplify this by expanding out these two terms and canceling these 3*x*s.

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

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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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

- Go to Continuity

- Go to Series

- Go to Limits

- Using Limits to Calculate the Derivative 8:11
- The Linear Properties of a Derivative 8:31
- Calculating Derivatives of Trigonometric Functions 7:20
- Calculating Derivatives of Polynomial Equations 10:25
- Calculating Derivatives of Exponential Equations 8:56
- Using the Chain Rule to Differentiate Complex Functions 9:40
- Differentiating Factored Polynomials: Product Rule and Expansion 6:44
- When to Use the Quotient Rule for Differentiation 7:54
- Calculating Higher Order Derivatives 9:24
- How to Find Derivatives of Implicit Functions 9:23
- How to Calculate Derivatives of Inverse Trigonometric Functions 7:48
- Applying the Rules of Differentiation to Calculate Derivatives 11:09
- Optimization Problems in Calculus: Examples & Explanation 10:45
- Go to Calculating Derivatives and Derivative Rules

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