# "Let f and g be twice differentiable functions ..." - Derivitive problem

• Mar 11th 2009, 03:36 PM
JTG2003
"Let f and g be twice differentiable functions ..." - Derivitive problem
There's a problem that our professor wanted us to look at and I'd really like to understand it. It says

"Let f and g be twice differentiable functions. Derive a formula for (fg'') = [(fg)']' using your knowledge of differentiation rules."

We know about the power rule and chain rule and all that fun stuff .. I guess I'm not sure what she's asking.

Am I trying to apply these rules to "[(fg)']'" to make it look like (fg'')? If I am, I don't know how I would even start that.

Any help would be appreciated.

Thanks
• Mar 11th 2009, 04:36 PM
Scott H

\$\displaystyle (fg)''=[(fg)']'=(f'g+g'f)'\$

by the Product Rule. By linearity,

\$\displaystyle (f'g+g'f)'=(f'g)'+(g'f)'.\$
• Mar 11th 2009, 04:37 PM
skeeter
Quote:

Originally Posted by JTG2003
There's a problem that our professor wanted us to look at and I'd really like to understand it. It says

"Let f and g be twice differentiable functions. Derive a formula for (fg'') = [(fg)']' using your knowledge of differentiation rules."

We know about the power rule and chain rule and all that fun stuff .. I guess I'm not sure what she's asking.

Am I trying to apply these rules to "[(fg)']'" to make it look like (fg'')? If I am, I don't know how I would even start that.

Any help would be appreciated.

Thanks

your notation should be \$\displaystyle (fg)'' = [(fg)']'\$

\$\displaystyle (fg)' = fg' + f'g\$

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

so ...

\$\displaystyle (fg)'' = [(fg)']' = fg'' + 2(f'g') + f''g\$
• Mar 11th 2009, 05:13 PM
JTG2003
Hm... ok, I think I get it.

It will take a few times of going through it I guess.

Thank you.