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?
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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) $
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) $
Originally Posted by xterminal01 Where does the +2x arctan(x) come from? the derivative of the product. $\displaystyle (f\times g)'=f'\times g+f\times g$'.
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, $\displaystyle (\arctan(x))'=\frac{1}{1+x^2}$
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