# Thread: Finding derivative of function

1. ## 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?

2. 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.
$\displaystyle f'(x)=x^2/(x^2+1)+2x \arctan(x)$

3. Where does the +2x arctan(x) come from?

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

4. Originally Posted by xterminal01
Where does the +2x arctan(x) come from?
the derivative of the product.
$\displaystyle (f\times g)'=f'\times g+f\times g$'.

5. Originally Posted by Raoh
the derivative of the product.
$\displaystyle (f\times g)'=f'\times g+f\times g$'.
anyway i think u have a problem,
$\displaystyle (\arctan(x))'=\frac{1}{1+x^2}$