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amyna

  • one year ago

the directions to this problem say: "differentiate" u=(arcsinx)^2 i dont understand what the "u" means? Do i just take the derivative as normal by using the chain rule by witting u' ??

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  1. jim_thompson5910
    • one year ago
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    I think they want you to derive with respect to x. So think of the `u` as like `y` ie you have y = (arcsin(x))^2

  2. zepdrix
    • one year ago
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    Yes, you have to find u' :)

  3. amyna
    • one year ago
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    ok thank guys!

  4. amyna
    • one year ago
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    oh ya and for this problem, how do i use the chain rule? f(x)=ln(arctanx)?

  5. amyna
    • one year ago
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    alright thanks!

  6. zepdrix
    • one year ago
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    \[\large\rm f(x)=\ln(\arctan x)\]\[\large\rm f'(x)=\frac{1}{\arctan x}(\arctan x)'\]Yes, chain rule :D Sorry, had to correct my mistake from before lol

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