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### Course: AP®︎ Calculus AB (2017 edition) > Unit 3

Lesson 13: Optional videos- Justifying the power rule
- Proof of power rule for positive integer powers
- Proof of power rule for square root function
- Limit of sin(x)/x as x approaches 0
- Limit of (1-cos(x))/x as x approaches 0
- Proof of the derivative of sin(x)
- Proof of the derivative of cos(x)
- Product rule proof
- Proof: Differentiability implies continuity
- If function u is continuous at x, then Δu→0 as Δx→0
- Chain rule proof
- Quotient rule from product & chain rules

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# 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?

- 1:40How 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)

- At2:40why 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.