# Finding derivative of function

• Dec 18th 2009, 01:40 PM
xterminal01
Finding derivative of function
f(x) = $\displaystyle x^2 arctan(x)$

$\displaystyle arctan(x) => X/(1+x^2)$
Then..
$\displaystyle x^2(1/1+(x)^2)$
$\displaystyle = x^2/1+x^2$
Is this done correctly?
• Dec 18th 2009, 01:44 PM
Raoh
Quote:

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

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

no,that's not right.
$\displaystyle f'(x)=x^2/(x^2+1)+2x \arctan(x)$
• Dec 18th 2009, 01:47 PM
xterminal01
Where does the +2x arctan(x) come from?

Quote:

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

• Dec 18th 2009, 01:51 PM
Raoh
Quote:

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

the derivative of the product(Nod).
$\displaystyle (f\times g)'=f'\times g+f\times g$'.
• Dec 18th 2009, 01:53 PM
Raoh
Quote:

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

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