## AJW99 Group Title 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? one year ago one year ago

1. Jemurray3 Group Title

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

2. AJW99 Group Title

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

3. AJW99 Group Title

(8x)^2 doesn't work either

4. Jemurray3 Group Title

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. Jemurray3 Group Title

on*

6. AJW99 Group Title

Thanks!