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anonymous

  • 5 years ago

Can someone check my Ln derivitve?

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  1. anonymous
    • 5 years ago
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    post it

  2. anonymous
    • 5 years ago
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    \[f(x)=\ln(\sin^2*x)\]

  3. anonymous
    • 5 years ago
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    that is sin^2(x) right?

  4. anonymous
    • 5 years ago
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    f'(x)=ln(sin^2*x)+1/2cosin1

  5. anonymous
    • 5 years ago
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    thats correct

  6. anonymous
    • 5 years ago
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    ((cos(x)*2sin(x))/(sin^2(x))

  7. anonymous
    • 5 years ago
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    that 2nd equation is a bit sloppy

  8. anonymous
    • 5 years ago
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    ok thanks

  9. anonymous
    • 5 years ago
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    am i right? i'm not even sure lol

  10. anonymous
    • 5 years ago
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    yep i'm right... you can simplify it to just 2cotx

  11. anonymous
    • 5 years ago
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    well i am thinking that because of the properties of natural log we use the coefficent of the ln for the numerator and whatever follows the ln as the denomator. so becaues we just have ln(sin^2x) we then get ((1)/2cos(x))

  12. anonymous
    • 5 years ago
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    But i am probably wrong

  13. anonymous
    • 5 years ago
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    why would you do that though? I guess you can simplify it by using 2ln(sin(x)), and then just move the 2 out. which would get the same answer..., because ln (sinx) = cos(x)/sin(x) = cot(x) *2

  14. anonymous
    • 5 years ago
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    the derivative of ln is 1/x btw.

  15. anonymous
    • 5 years ago
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    ok but we wouldn't replace the x with whatever we are apply the natural log to? doesn't ln(x)^5 = 5lnx?

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