## anonymous 3 years ago Let f(x)=(x^2)tan^-1(8x) find f'(x) I know that the derivative of arctan is 1/(1+x^2) So, when I applied the product rule I got (x^2)(1/(1+8x^2))+(arctan(8x))(2x). But this isn't correct. Where am I going wrong on this problem?

1. anonymous

$(8x)^2 \neq 8x^2$

2. anonymous

If I switch it to be 8x^2 it still isn't correct

3. anonymous

(8x)^2 doesn't work either

4. anonymous

If $f(x) = x^2\tan^{-1}(8x)$ then $f'(x) = 2x\tan^{-1}(8x) + \frac{8x^2}{1+64x^2}$ You have to use the chain rule or the last part.

5. anonymous

on*

6. anonymous

Thanks!