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AJW99

  • 2 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?

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  1. Jemurray3
    • 2 years ago
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    \[(8x)^2 \neq 8x^2\]

  2. AJW99
    • 2 years ago
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    If I switch it to be 8x^2 it still isn't correct

  3. AJW99
    • 2 years ago
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    (8x)^2 doesn't work either

  4. Jemurray3
    • 2 years ago
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    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
    • 2 years ago
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    on*

  6. AJW99
    • 2 years ago
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    Thanks!

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