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### Course: AP®︎/College Calculus AB>Unit 3

Lesson 10: Optional videos

# Quotient rule from product & chain rules

We explore the connection between the quotient rule, product rule, and chain rule in calculus. Rather than memorizing another rule, we see how the quotient rule naturally emerges from applying the product and chain rules. This approach simplifies our understanding of these fundamental calculus tools. Created by Sal Khan.

## Want to join the conversation?

• How does g(x) becomes [g(x)]^-1?
(5 votes)
• Because we originally had f(x)/g(x) which is =f(x)*1/g(x)=f(x)*[g(x)]^-1,
that is why 1/g(x)=g(x)^-1, and actually this is true for any number x except zero.
So
1/x=x^-1
and similarly we can substitute in any value for x except zero.
1/2=2^-1
1/3=3^-1
1/777=777^-1
If you don't believe me, try it out using a calculator ;-)
Hope this helps :-)
(18 votes)
• I don't have a questions about this video but I am having trouble with problems that require you to take multiple types of derivatives..

For example:
y = tan [ ln (ax + b) ]
Do we have to use chain rule or is it just product?
(2 votes)
• Be careful - the only multiplication going on in that problem is the "ax" part. This is not a product rule problem.

This is a chain rule, within a chain rule problem. The rule remains the same, you just have to do it twice: differentiate the outermost function, keep the inside the same, then multiply by the derivative of the inside.

= sec^2[ ln (ax + b) ] * d/dx[ ln (ax + b]
= sec^2[ ln (ax + b) ] * (ax + b)^-1 * d/dx (ax + b)
= sec^2[ ln (ax + b) ] * (ax + b)^-1 * a
(7 votes)
• Is the quotient rule a combination of the product and chain rule, or was that just an example?
(2 votes)
• The quotient rule could be seen as an application of the product and chain rules. If Q(x) = f(x)/g(x), then Q(x) = f(x) * 1/(g(x)) . You can use the product rule to differentiate Q(x), and the 1/(g(x)) can be differentiated using chain rule with u = g(x), and 1/(g(x)) = 1/u. This is what Sal does in the video.
(6 votes)
• Why this " ´(f(x)/g(x)) = f ´(x) * 1/g(x) + f(x) * 1/g ´(x) " do not works?

I understand that Sal is Using the product rule and then the chain rule, but I just do not understand why using just the product rule do not works.

I can not explain my doubt better I hope someone could understand my concern and explain me why using the product rule and the chain rule to solve this works good but using the product rule do not.
(3 votes)
• This does not work because of the final term in your equation. In order for it to work you need to take d/dx(1/g(x)). You only took the derivative of g(x) instead of taking the derivative of the whole thing. Otherwise, you have the basic understanding down correctly
(3 votes)
• Does it matter whether we take the derivative of f(x) first or g(x) first in the numerator of the result?
(3 votes)
• If you are using the product rule, it will not matter. If you are using the quotient rule, it will matter. Another advantage of the product rule.
(2 votes)
• what can be the answer for F'(x) = log x and G'(x) = e^x?? what and how??
(2 votes)
• log(x) is in base a.
If F'(x)=log(x) and G'(x)=e^x, then F(x)=x*log(x)-x/ln(a) (plus any constant) and G(x)=e^x (plus any constant). I took the anti-derivative of each function, which is kind of hard to explain. Sal has a playlist on it.
https://www.khanacademy.org/math/calculus/integral-calculus/indefinite_integrals/

We'll call the constant we added to F(x) C and the constant we added to G(x) c.

F(x)=x*log(x)-x/ln(a)+C
G(x)=e^x+c
F'(x)=log(x)
G'(x)=e^x
Substitute.
(f'(x)g(x)-f(x)g'(x))/(g(x)^2)
(log(x)(e^x+c)-(x*log(x)-x/ln(a)+C)(e^x)/((e^x+c)^2)
It's probably best that we keep it in this form so it doesn't get crazy.

Thus, the derivative of the function F(x)/G(x) is
(log(x)(e^x+c)-(x*log(x)-x/ln(a)+C)(e^x)/((e^x+c)^2)
for all constants C and c.
I hope this helped!
(4 votes)
• At why Sal multiplies (g(x)^-1) derivative, by g`(x) again? please someone can explain me whats wrong with my thinking?
(2 votes)
• He is using the chain rule, first he takes the derivative of g(x))^-1 with respect to g(x), which is
-g(x)^-2;;; and then it has to be multiplied by the derivative of the inner function with respect to x, which is g'(x)
(3 votes)
• usually in functions, when we say [g(x)]^2, it is g(g(x)). for example, if g(x)=x+1, in the quotient rule, why is [g(x)]^2 =(x+1)^2 and not [(x+1)+1] =x+2?
(2 votes)
• I don't think there is anything usual about that at all.
I agree g²(x) could well mean g(g(x)), but I've never seen [g(x)]² mean anything other than g(x)·g(x).
I'm not saying your interpretation is impossible, just that it's not "usual".
(3 votes)
• Does it matter which way the f'(x) or g'(x) goes?
Cus i tried doing (ln x/cos x) and it kept putting after the f'(x) g(x) + it kept putting the f(x) which is ln x after the -sin x?
(2 votes)
• I think you are asking (please correct me if I am wrong) if the order counts for this formula, and the answer is yes. Because of the subtraction (which is order specific) you need to use the proper order.

[ f'(x)g(x) - f(x)g'(x) ] / (g(x))^2
(2 votes)
• Btw when do you know if you should plus or minus?
(2 votes)
• He used the power rule on [g(x)]^(-1). So taking down the (-1) makes the second term a negative
(2 votes)

## Video transcript

We already know that the product rule tells us that if we have the product of two functions-- so let's say f of x and g of x-- and we want to take the derivative of this business, that this is just going to be equal to the derivative of the first function, f prime of x, times the second function, times g of x, plus the first function, so not even taking its derivative, so plus f of x times the derivative of the second function. So two terms, in each term we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit faster, but it really comes straight out of the product rule. And I frankly always forget the quotient rule, and I just rederive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business, the derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of-- instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule. It's the derivative of the first function right over here-- so it's going to be f prime of x-- times just the second function, which is just g of x to the negative 1 power plus the first function, which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of view as something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your math book. So let's see if we can simplify this a little bit. All of this is going to be equal to-- we can write this term right over here as f prime of x over g of x. And we could write all of this as-- we could put this negative sign out front. We have negative f of x times g prime of x. And then all of that over g of x squared. Let me write this a little bit neater. All of that over g of x squared. And it still isn't in the form that you typically see in your calculus book. To do that, we just have to add these two fractions. So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator. So if we multiply the numerator by g of x, we'll get g of x right over here and then the denominator will be g of x squared. And now we're ready to add. And so we get the derivative of f of x over g of x is equal to the derivative of f of x times g of x minus-- not plus anymore-- let me write it in white-- f of x times g prime of x, all of that over g of x squared. So once again, you can always derive this from the product rule and the chain rule. Sometimes this might be convenient to remember in order to work through some problems of this form a little bit faster. And if you wanted to kind of see the pattern between the product rule and the quotient rule, the derivative of one function just times the other function. And instead of adding the derivative of the second function times the first function, we now subtract it. And all that is over the second function squared. Whatever was in the denominator, it's all of that squared. So when we're taking the derivative of the function in the denominator up here, there's a subtraction, and then we are also putting everything over the second function squared.