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Quotient rule

# Quotient rule

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.B (LO)
,
FUN‑3.B.2 (EK)
Introduction to the quotient rule, which tells us how to take the derivative of a quotient of functions.

## Want to join the conversation?

• Why is the derivative of x^2 equal to 2x?
(1 vote)
• In this video, he gives the example of (x^2)/cos(x). But wouldn't there be an asymptote at x=pi/2? Why is it we can differentiate this function if it is not continuous?
(3 votes)
• The function is continuous and differentiable everywhere except where cos(x)=0. The function has vertical asymptotes, and its derivative has asymptotes in the same places.
(6 votes)
• I think I'm a little shaking on simplifying some of these expressions. Could somebody please help me out...
d/dx √x/cos(x) =1/2√x · cos(x) - √x · -sin(x)/cos²(x)
=cos(x)+√x sin(x) / 2√x · cos²(x)
Is that right so far? The final expression = cos(x)+2xsin(x) / 2√x·cos²(x), as per Kahn Academy.
How did they get to 2xsin(x) in the numerator? I must be missing something.
(3 votes)
• Can someone explain to me why we sometimes can differentiate simple functions using the power rule (such as f(x) = 1/(x^2) = x^-2, thus f'(x) = -2(x)^-3) and sometimes we require the quotient rule to solve these problems? [For example, g(x) = 1/(x^2 - 1) which would leat to f'(x) = -(2x)/(x^4-1)?] Thank you!
(1 vote)
• We can always use the power rule instead of the quotient rule. However, this isn't possible without another rule called the chain rule, so it's best to stick with the quotient rule until you learn the chain rule.

On another note, I believe you may have made a mistake in your use of the quotient rule for your g(x) function. The correct answer for g'(x) should be (x^2-2x-1)/(x^4-2x+1). I think you may have made a mistake by cancelling the (x^2-1) in the denominator with the one in the numerator. You can't do that because the one in the numerator also has a 2x being subtracted, so there aren't actually common factors to cancel.
(5 votes)
• I'm beginning to think f is a lie.
(3 votes)
• I noticed that a proof is not available in this section of derivatives. Can someone explain to me the proof?
(2 votes)
• At I'm not seeing how the X^2 becomes positive
(3 votes)
• Review the algebra/precalculus playlists.
(0 votes)
• My teacher taught us this except he said to do : (Low)(D-High) - (High)(D-Low) and then the bottom fraction squared. You have the same setup except the derivatives are switched and you have the top fraction coming first in the equation instead of the bottom one. I have a quiz on this tomorrow and now I'm just confused.
(2 votes)
• This is no difference. The order of the terms ((low)(D-high)) changes. Multiplication of numbers’ orders can be switched.
(1 vote)
• Does it matter if I write (f)(g')-(f')(g) instead of (f')(g)-(f)(g')
(1 vote)
• Yes. You cannon switch terms in subtraction. That would result in the opposite(negative) of the previous expression. However, -(f'(x)*g(x)-f(x)*g'(x)) does equal f(x)*g'(x)-f'(x)*g(x), by the distributive property.
(1 vote)
• On the product rule video, I commented a way to memorize the rule, then went on to say I had a way to memorize the quotient rule. Just say “f’g-g’f/g^2” Or, the more confusing but more fun, in my opinion, “Low dee high minus high dee low, square the low there you go” translatestion: g(x)d/dx(f(x))-f(x)d/dx(g(x))/g^2(x)
(2 votes)

## Video transcript

- [Instructor] What we're going to do in this video is introduce ourselves to the quotient rule. And we're not going to prove it in this video. In a future video we can prove it using the product rule and we'll see it has some similarities to the product rule. But here, we'll learn about what it is and how and where to actually apply it. So for example if I have some function F of X and it can be expressed as the quotient of two expressions. So let's say U of X over V of X. Then the quotient rule tells us that F prime of X is going to be equal to and this is going to look a little bit complicated but once we apply it, you'll hopefully get a little bit more comfortable with it. Its going to be equal to the derivative of the numerator function. U prime of X. Times the denominator function. V of X. Minus the numerator function. U of X. Do that in that blue color. U of X. Times the derivative of the denominator function times V prime of X. And this already looks very similar to the product rule. If this was U of X times V of X then this is what we would get if we took the derivative this was a plus sign. But this is here, a minus sign. But were not done yet. We would then divide by the denominator function squared. V of X squared. So let's actually apply this idea. So let's say that we have F of X is equal to X squared over cosine of X. Well what could be our U of X and what could be our V of X? Well, our U of X could be our X squared. So that is U of X and U prime of X would be equal to two X. And then this could be our V of X. So this is V of X. And V prime of X. The derivative of cosine of X with respect to X is equal to negative sine of X. And then we just apply this. So based on that F prime of X is going to be equal to the derivative of the numerator function that's two X, right over here, that's that there. So it's gonna be two X times the denominator function. V of X is just cosine of X times cosine of X. Minus the numerator function which is just X squared. X squared. Times the derivative of the denominator function. The derivative of cosine of X is negative sine X. So, negative sine of X. All of that over all of that over the denominator function squared. So that's cosine of X and I'm going to square it. I could write it, of course, like this. Actually, let me write it like that just to make it a little bit clearer. And at this point, we just have to simplify. This is going to be equal to let's see, we're gonna get two X times cosine of X. Two X cosine of X. Negative times a negative is a positive. Plus, X squared X squared times sine of X. Sine of X. All of that over cosine of X squared. Which I could write like this, as well. And we're done. You could try to simplify it, in fact, there's not an obvious way to simplify this any further. Now what you'll see in the future you might already know something called the chain rule, or you might learn it in the future. But you could also do the quotient rule using the product and the chain rule that you might learn in the future. But if you don't know the chain rule yet, this is fairly useful.