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Course: Calculus, all content (2017 edition)>Unit 2

Lesson 22: Quotient rule

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.

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• How does g(x) becomes [g(x)]^-1?
• 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 :-)
• 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?
• 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
• Is the quotient rule a combination of the product and chain rule, or was that just an example?
• 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.
• 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.
• 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
• Does it matter whether we take the derivative of f(x) first or g(x) first in the numerator of the result?
• 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.
• what can be the answer for F'(x) = log x and G'(x) = e^x?? what and how??
• 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.

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!
• At why Sal multiplies (g(x)^-1) derivative, by g`(x) again? please someone can explain me whats wrong with my thinking?
• 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)
• 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?
• 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".
• 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?