anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[(8x)^2 \neq 8x^2\]
anonymous
  • anonymous
If I switch it to be 8x^2 it still isn't correct
anonymous
  • anonymous
(8x)^2 doesn't work either

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anonymous
  • 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.
anonymous
  • anonymous
on*
anonymous
  • anonymous
Thanks!

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