# differentiating a function...

• Jan 29th 2009, 09:00 AM
bemidjibasser
differentiating a function...
I think I have this, but want to make sure...
Let g(x)= x^2 f(x)
What does g'(x)=?
Using the power rule I come up with (2x * f(x) * f'(x) )*(x^2 * f'(x))
Is that right?
Any help would be greatly appreciated.
• Jan 29th 2009, 09:02 AM
Moo
Hello,
Quote:

Originally Posted by bemidjibasser
I think I have this, but want to make sure...
Let g(x)= x^2 f(x)
What does g'(x)=?
Using the power rule I come up with (2x * f(x) * f'(x) )*(x^2 * f'(x))
Is that right?
Any help would be greatly appreciated.

Nope, you have to use the product rule :
\$\displaystyle [uv]'=u'v+uv'\$, where \$\displaystyle u=x^2\$ and \$\displaystyle v=f\$
• Jan 29th 2009, 09:03 AM
bemidjibasser
moo...
could you refresh me on how to work through that?
• Jan 29th 2009, 09:08 AM
Moo
Quote:

Originally Posted by bemidjibasser
could you refresh me on how to work through that?

\$\displaystyle g(x)=x^2 *f(x)\$

The product rule of differentiation says that
\$\displaystyle g'(x)=[x^2]'*f(x)+x^2 *[f(x)]'=2x*f(x)+x^2*f'(x)\$

Does it look clear ?
• Jan 29th 2009, 09:11 AM
bemidjibasser
yeah, that's what i had, right?
• Jan 29th 2009, 09:16 AM
Moo
Quote:

Originally Posted by bemidjibasser
yeah, that's what i had, right?

Quote:

(2x * f(x) * f'(x) )*(x^2 * f'(x))
Quote:

2x*f(x)+x^2*f'(x)
This is not the same at all (Surprised)
• Jan 29th 2009, 09:33 AM
bemidjibasser
ooops...
I had a * not a + thanks a bunch MOO!