Question

True/False: Derivatives: Let f(x) and g(x) be two functions for which the 2nd derivatives are defined. Then (fg)''=f''g + fg'' .

I was thinking true, because of the basic product rule for derivatives. Can't even begin to explain why it would be false, but I was thinking it would be longer... something along the lines of f'g + fg' + f'g + fg' . However since the 2nd derivative is defined, I'm stuck between the T/F. Thanks for any help/comments.

Answer 1

We need to find
\[(fg)''
\]
We know by product rule of differentiation
\[(uv)'=uv'+u'v\]
First let's find (fg)'
\[(fg)'=f'g+g'f\]
Now
\[(fg)''=((fg)')'=(f'g+g'f)'=(f'g)'+(g'f)'\]
or
\[=f''g+f'g'+g''f+g'f'=>f"g+2f'g'+g"f\]

Answer 2

\((fg)'=f'g + fg' \) via chain rule

\[(fg)''=(f'g + fg' )' = f''g+f'g' + fg''+f'g'\]\[=f''g+2f'g'+fg'' \neq f''g + fg'' \]

Answer 3

Ah so it was much longer, thanks.

Answer 4

The rule is Product rule not chain rule, My apologies.

Answer 5

Apology accepted :P