Question

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?

Answer 1

\[(8x)^2 \neq 8x^2\]

Answer 2

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

Answer 3

(8x)^2 doesn't work either

Answer 4

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.

Answer 5

on*

Answer 6

Thanks!