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.

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TopNewestWow! – Kulkul Chatterjee · 3 years, 3 months ago

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Hint: \(f(x) = a(x), \; g(x) = b(x) - a(x)\)? – Guilherme Dela Corte · 3 years, 3 months ago

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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. – Hussein Hijazi · 3 years ago

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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 \(g\). – Calvin Lin Staff · 3 years ago

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– Hussein Hijazi · 3 years ago

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

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. – Hussein Hijazi · 3 years ago

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– Calvin Lin Staff · 3 years ago

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

– Hussein Hijazi · 3 years ago

My pleasure!Log in to reply