# $(f \circ g) ' = f' \circ g + f \circ g '$

Sometimes, we can confuse applying the chain rule with applying the product rule.

The product rule states that $[ f(x) \times g(x) ]' = f'(x) \times g(x) + f(x) \times g'(x),$ while the chain rule states that $(f \circ g )' (x) = g'(x) \times f' \circ g (x)$.

Find infinitely many pairs of functions such that

$(f \circ g) ' = f' \circ g + f \circ g '$

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.

Note by Calvin Lin
6 years, 4 months ago

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Wow!

- 6 years, 3 months ago

Hint: $f(x) = a(x), \; g(x) = b(x) - a(x)$?

- 6 years, 4 months ago

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.

- 6 years, 1 month ago

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 $g$.

Staff - 6 years, 1 month ago

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.

- 6 years, 1 month ago

I completely missed that. Thanks for pointing it out! I've made the corresponding edits.

Staff - 6 years, 1 month ago

My pleasure!

- 6 years, 1 month ago

Unless what you stated is the perceived "confused" version, then my bad. 'Twas a bit hard to notice

- 6 years, 1 month ago

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