Sometimes, we can confuse applying the chain rule with applying the product rule.
The product rule states that [f(x)×g(x)]′=f′(x)×g(x)+f(x)×g′(x), while the chain rule states that (f∘g)′(x)=g′(x)×f′∘g(x).
Find infinitely many pairs of functions such that
(f∘g)′=f′∘g+f∘g′
Example: f(x)=e−x and g(x)=1+x−e−x.
This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.
#Calculus
#Proofs
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Wow!
Hint: f(x)=a(x),g(x)=b(x)−a(x)?
I'm pretty sure you have the rules confused. The first one you listed is the product rule, as it is the multiplcation of two functions. The second one, which has f and g as composite functions, is in fact the chain rule...
Pretty sure thats what going on.
Log in to reply
As stated, "Sometimes, we can confuse applying the chain rule with applying the product rule."
I am intentionally asking you to find pairs of functions in which the "confused" version ends up being correct. It is certainly not true of any (differentiable) functions f and g.
Log in to reply
I understand that. But in the introduction where you say "The chain rule states that [Latex stuff] while the product rule states that [more Latex stuff]" but those Latex areas should be switched because the chain rule does not state what you said nor does the product rule state what is followed.
I understand some composite functions, when differentiated, turn out to be like the product rule. But what you said in the beginning isn't true.
If it's some kind of humor I'm not seeing, I'm sorry.
Log in to reply
Log in to reply
Unless what you stated is the perceived "confused" version, then my bad. 'Twas a bit hard to notice