im lost...
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Why don't you try a little by yourself ? You may know the derivate of a product. The derivate of arctan(u(x)) is u'(x)/(1+uČ(x))
The first thing we have is a product rule. There is also one chain rule. So: $\displaystyle f'(x) = 6x^3\cdot \frac {1}{1+81x^4}\cdot 18x + 18x^2\arctan {9x^2}$ Then just simplify.
$\displaystyle f(x)=6x^3 \tan^{-1}(9x^2)$ $\displaystyle f'(x)=18x^2 \tan^{-1}(9x^2)+6x^3 \cdot \left( \frac{18x}{1+(9x^2)^2} \right)$ Just simplify from here. Good luck.
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