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### Course: AP®︎/College Calculus AB > Unit 2

Lesson 10: The quotient rule# Differentiating rational functions

Let's dive into the differentiation of the rational function (5-3x)/(x²+3x) using the Quotient Rule. By identifying the numerator and denominator as separate functions, we apply the Quotient Rule to find the derivative, simplifying the expression for a clear understanding of the process. This approach can be applied to differentiate any other rational function as well.

## Want to join the conversation?

- Why does sal never simplify the
`((x^2)+3x)^2`

in the denominator?(16 votes)- Because simplifying the denominator can lead to unnecessary algebraic mistakes and a more cluttered fraction. My calculus teacher doesn't even have us simplify our answers to problems because even the smallest mistake in solving for the answer can lead to an incorrect answer. Leaving the denominator in this form ensures the mathematician that they don't have to worry about small mistakes(44 votes)

- What does it mean for the derivative to exist at every point on the domain of it's function? It seems like an abstract thing to visualize.

Can someone explain?(6 votes)- It means that for all real numbers (in the domain) the function has a derivative.

For this to be true the function must be defined, continuous and**differentiable**at all points.

In other words, there are no discontinuities, no corners AND no**vertical tangents**.

ADDENDUM: An example of the importance of the last condition is the function`f(x) = x^(1/3)`

— this function is defined and continuous for all real numbers, but at`x=0`

it has a vertical tangent and thus is not differentiable.(6 votes)

- Why would you need to do partial fraction if you have quotient rule?(4 votes)
- You wouldn't, for differentiation. However, for solving some integrals, partial fraction decomposition is the only way to make progress.(4 votes)

- In this lesson Sal uses dy/dx where he only used and explained d/dx in the past. What is the difference between the two, if any?(3 votes)
- d/dx - Derivative of [blank] with respect to x. In and of itself, it's meaningless, like a plus sign operator by itself.

dy/dx - Derivative of y with respect to x. An actual quantity or function (if y is differentiable).(5 votes)

- take d/dx[(2x + 3)/(3x^2 - 4)] at x=-1. Say f(x) = 2x + 3 and g(x) = 3x^2 - 4. So that's [f'(x)g(x) - f(x)g'(x)] / [g(x)]^2.

Plug in the expressions to get d/dx(2x + 3)(3x^2 -4) - (2x+3)d/dx(3x^2 - 4), all divided by (3x^2 - 4)^2.

If you do the math, the derivative is 4.

But what happens if, when taking the derivative of a constant, you made it 1 instead of 0? So d/dx(2x + 3) = 2 + 1 = 3.

Do that through the whole problem. The final answer is still 4. And this is true of not just this expression. How is this happening?(2 votes)- The derivative of a constant
**is**0 not 1.

If you are getting the same answer using both methods that is just a coincidence — it is not universally true.

I suggest trying this with more examples — you will quickly find that it usually does**not**work!(3 votes)

- I rewrote f(x) as (5 - 3x)*(x^2 + 3x)^(-1) and used the product rule. The answer was different. Why can't I rewrite the function to be a product and then find its derivative? Dividing by something is the same as multiplying by its reciprocal(2 votes)
- You must have made an error in your math somewhere. You should get the same answer when you use the product rule.

f(x)=(5-3x)*(x^2+3x)^(-1)

f'(x)=(5-3x)'*(x^2+3x)^(-1)+(5-3x)*((x^2+3x)^(-1))'

f'(x)=-3*(x^2+3x)^(-1)-(5-3x)*(2x+3)*(x^2+3x)^(-2)

f'(x)=(-3*(x^2+3x)-(5-3x)(2x+3))*(x^2+3x)^(-2)

f'(x)=(-3*(x^2+3x)-(5-3x)(2x+3))/(x^2+3x)^2

f'(x)=(-3x^2-9x-10x-15+6x^2+9x)/(x^2+3x)^2

f'(x)=(3x^2-10x-15)/(x^2+3x)^2(2 votes)

- Did Mr.Khan just use the FOIL method4:20? Gasp(1 vote)
- In one of the test questions, I got the wrong answer because I didn't expand the brackets and I'm confused why one is correct and the other is not. Basically, the question asked for the derivative at x=1 and for simplicity let's say the function was just u/v. Well u' evaluated to (2x-2) = 0. Therefore I didn't need to care about the u'v part of the quotient rule. And my answer was (-(v' * u))/v^2. But in Khan's answer, all the brackets were expanded first so that there were no zeros.(1 vote)
- Would you happen to have the exact question? The way you put it, I don't see any error. But, maybe if I have the question, I can pinpoint the problem.(1 vote)

## Video transcript

- [Voiceover] Let's say that y is equal to five minus three
x over x squared plus three x And we want to figure out what's the derivative
of y in respect to x. Now it might immediately
jump out at you that look, y is being defined as a
rational expression here as the quotient of two
different expressions. We could even view this as
two different functions. You could view this one up here as u of x, so you could say this is the same thing, this is the same thing as u of x, over, you could view the one in
the denominator as v of x. That one right there is v of x and so if you're taking
the derivative of something that can be expressed in this way, as the quotient of two
different functions, well then you could use the quotient rule. I'll give you my little
aside, like I always do the quotient rule, if you ever forget it it can be derived from the product rule and we have videos there, cause the product rule's
a bit easier to remember. But what I can do is just say, "look, dy dx, if y is
just u of x over v of x", I'm just gonna restate the quotient rule. This is going to be, this is going to be the derivative of the
function in the numerator, so. d, dx of u of x times the function in the denominator times v of x minus, I'll do the minus the function in
the numerator, u of x, times the derivative of the
function in the denominator times d dx, v of x and we're almost there and then over, over the function
in the denominator squared the function in the denominator squared. So this might look messy
but all we have to do now is think about what is
the derivative of u of x? What is the derivative of v of x? And we should just be able
to substitute those things back into this expression
we just wrote down. So let's do that. So the derivative with respect
to x of u of x, of u of x is equal to, let's see,
five minus three x. The derivative of five is zero. The derivative of negative three x, well that's just gonna be negative three. That's just negative three. If any of that look
completely unfamiliar to you I encourage you to review
the derivative properties and maybe the power rule. Now let's think about
what is the derivative with respect to x,
derivative with respect to x, of v of x, of v of x? Well derivative of x squared, we just bring that exponent out front, it's gonna be two times
x to the two minus one or two x to the first power or just two x and then the derivative
of three x is just three. So two x plus three. Now we know everything we need
to substitute back in here. The derivative of u with respect to x, this right over here
is just negative three. V of x, this we know is
x squared plus three x. We know that this right
over here is v of x. And then u of x we know
is five minus three x. Five minus three x. The derivative of v with respect to x we know is two x plus
three, two x plus three. And then finally v of x, we
know is x squared plus three x. So this is x squared plus three x and so what do we get? Well we are going to get, it's
gonna look a little bit hairy It's going to be equal to negative I'll focus this so first we
have this business up here. Negative three times x
squared plus three x. So I'm just gonna distribute
the negative three. So it's negative three
x squared minus nine x and then from that we
are going to subtract the product of these two
expressions and so let's see, what is that going to be? Well, we have a five times two x, which is ten x a five times three which is 15 we have a negative three x times two x so that is going to be
negative six x squared minus six x squared, and then
a negative three x times three so negative nine x. And let's see we can
simplify that a little bit. Ten x minus nine x, well that's just going to leave us with x. So ten x minus nine x
is just going to be x and then in our denominator,
we're almost there. In our denominator we
could just write that as x plus three x, x squared
plus three x squared or if we want we could expand it out, I'll just leave it like that x squared plus three x squared and so if we wanna simplify or attempt to simplify this a little bit. It's going to be negative
three x squared minus nine x and then you're gonna
have a negative minus x minus x and then minus 15 and then minus negative six x squared so plus six x squared, all of that over x squared plus three x squared or x squared plus three x, squared. I should say it that way. Now let's see this numerator
I can simplify a little bit. Negative three x squared
plus six x squared that's going to be
positive three x squared and then we have in orange, we have negative nine x minus an x, well that's gonna be
minus ten x, minus ten x and then we have minus 15. So minus 15. So there you have it, we
finally have finished. This is all going to be equal to this is all going to be equal to three x squared minus ten x minus 15 over x squared plus three x, squared and we are done.