# Finding derivative of function

• Dec 18th 2009, 02:40 PM
xterminal01
Finding derivative of function
f(x) = $x^2 arctan(x)$

$arctan(x) => X/(1+x^2)$
Then..
$x^2(1/1+(x)^2)$
$= x^2/1+x^2$
Is this done correctly?
• Dec 18th 2009, 02:44 PM
Raoh
Quote:

Originally Posted by xterminal01
f(x) = $x^2 arctan(x)$

$arctan(x) => X/(1+x^2)$
Then..
$x^2(1/1+(x)^2)$
$= x^2/1+x^2$
Is this done correctly?

no,that's not right.
$f'(x)=x^2/(x^2+1)+2x \arctan(x)$
• Dec 18th 2009, 02:47 PM
xterminal01
Where does the +2x arctan(x) come from?

Quote:

Originally Posted by Raoh
no,that's not right.
$f'(x)=x^2/(x^2+1)+2x \arctan(x)$

• Dec 18th 2009, 02:51 PM
Raoh
Quote:

Originally Posted by xterminal01
Where does the +2x arctan(x) come from?

the derivative of the product(Nod).
$(f\times g)'=f'\times g+f\times g$'.
• Dec 18th 2009, 02:53 PM
Raoh
Quote:

Originally Posted by Raoh
the derivative of the product(Nod).
$(f\times g)'=f'\times g+f\times g$'.

anyway i think u have a problem,
$(\arctan(x))'=\frac{1}{1+x^2}$
(Nod)