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Let f and g be two functions whose second derivatives are defined. Then (f · g) '' = f · g '' + f '' · g?
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\[\large{(f \cdot g) '' \\ = {d\over dx}\left( (f \cdot g)' \right) \\ = {d\over dx}(f'g + fg') \\ = {d\over dx}(f'g) + {d\over dx}(fg') \\ = (f')'(g) +(f')(g)' \ \ \ \ \ + \ \ \ \ \ (f)'(g') + (f)(g')' \\ = f''g + 2f'g' + fg'' }\]
2f'g' is probably not 0, so the statement is false.
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