Finding derivative of function

**xterminal01** · December 18th 2009, 01:40 PM

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?

**Raoh** · December 18th 2009, 01:44 PM

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

**xterminal01** · December 18th 2009, 01:47 PM

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

Originally Posted by Raoh

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

**Raoh** · December 18th 2009, 01:51 PM

Originally Posted by xterminal01

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

the derivative of the product

.
$(f\times g)'=f'\times g+f\times g$ '.

**Raoh** · December 18th 2009, 01:53 PM

Originally Posted by Raoh

the derivative of the product

.
$(f\times g)'=f'\times g+f\times g$ '.

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

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