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anonymous

  • one year ago

How can I find the derivative of the following function? f(x) = 5arctan(xe^x)

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  1. jim_thompson5910
    • one year ago
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    if y = arctan(x), then dy/dx = 1/(1+x^2)

  2. jim_thompson5910
    • one year ago
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    you will use the chain rule first, then the product rule will be used later on

  3. jim_thompson5910
    • one year ago
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    let me know if that helps or not to get started

  4. anonymous
    • one year ago
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    @jim_thompson5910 how do you get 1/(1+x^2) ? Is that just something i have to memorize? For example like how the derivative of sinx = cosx ?

  5. rational
    • one year ago
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    you may memorize but derivation isn't that hard

  6. rational
    • one year ago
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    \[\arctan x = y \implies x = \tan y \implies \frac{dx}{dy}=\sec^2y \] \[\implies \frac{dy}{dx}=\cos^2y = \frac{1}{1+x^2}\]

  7. anonymous
    • one year ago
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    Oh so i can use a triangle to find the derivative, correct? Thats what you did above right?

  8. rational
    • one year ago
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    Exactly!

  9. anonymous
    • one year ago
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    Okay I understand! I'm gonna try to derive this given what I understood, and I'll tell you my answer :D i feel like its gonna be wrong though xD

  10. rational
    • one year ago
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    ok sure good luck!

  11. anonymous
    • one year ago
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    Ok so this is what I did... \[\tan(y) = xe^x\] \[\frac{ dy }{ dx }\tan(y) = \frac{ dy }{ dx } xe^x\] \[\sec^2 yy' = e^x +xe^x\] \[y'=\frac{ e^x + xe^x }{ \sec^2 y } = e^x \frac{ 1+x }{ \sqrt{1+xe ^{2x}} }\] \[f'(x) = 5e^x \frac{ 1+x }{ \sqrt{1+xe ^{2x}} }\]

  12. anonymous
    • one year ago
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    Is that right? I can simplify it more right?

  13. anonymous
    • one year ago
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    I can't think of a way to simplify the denominator

  14. amilapsn
    • one year ago
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    please check what you've done for \(sec^2y\)...

  15. anonymous
    • one year ago
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    Ohhh I forgot to square it!! Right??

  16. amilapsn
    • one year ago
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    hm

  17. anonymous
    • one year ago
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    So then it would be \[y'=5 \frac{ e^x(1+x) }{ 1+2xe ^{2x}+x^2e ^{4x} }\]

  18. amilapsn
    • one year ago
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    no

  19. anonymous
    • one year ago
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    Oh no no no wait

  20. anonymous
    • one year ago
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    \[y'=5e^x \frac{ 1+x }{ 1+xe ^{2x} }\]

  21. amilapsn
    • one year ago
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    :o)

  22. anonymous
    • one year ago
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    Hahaha right?? XD

  23. anonymous
    • one year ago
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    But then what can I do next? Can I simplify it any more?

  24. anonymous
    • one year ago
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    Nope I'm still wrong!

  25. amilapsn
    • one year ago
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    hm

  26. amilapsn
    • one year ago
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    i didn't notice it..

  27. anonymous
    • one year ago
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    The part when I used the Pythagorean theorem to get \[\sqrt{1+xe ^{2x}}\] was wrong! It should be \[\sqrt{1+x^2e ^{2x}}\]

  28. amilapsn
    • one year ago
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    hm

  29. anonymous
    • one year ago
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    So at the end I get \[y'=5e^x \frac{ 1+x }{ 1+x^2e ^{2x} }\]

  30. anonymous
    • one year ago
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    Cuz \[(xe^x)^2 = x^2e ^{2x}\] and not \[xe ^{2x} \] is that right?

  31. amilapsn
    • one year ago
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    right.

  32. anonymous
    • one year ago
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    Awesome!!! Phewww haha that was such a struggle!! Thanks so so much @amilapsn and @rational

  33. amilapsn
    • one year ago
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    u're welcome.

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