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ammu123

  • 2 years ago

find the derivative of y with respect to x, t or theta as appropriate. y= ln(cos(ln theta))

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  1. fanitariq
    • 2 years ago
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    its very simple

  2. ammu123
    • 2 years ago
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    please help me

  3. fanitariq
    • 2 years ago
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    -(1/cos(ln theta))*(sin(ln theta)*1/theta

  4. stamp
    • 2 years ago
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    \[\frac{d(ln(cos(ln(\theta)))}{dx}\]

  5. stamp
    • 2 years ago
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    Should be dTheta

  6. ammu123
    • 2 years ago
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    the answer is - (tan (ln theta))/ theta

  7. ammu123
    • 2 years ago
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    i just dont no how to get to that

  8. stamp
    • 2 years ago
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    \[\frac{1}{cos(ln(\theta))}*-sin(ln(\theta))*\frac{1}{\theta}\]\[\frac{-sin(ln(\theta))}{cos(ln(\theta))}\frac{1}{\theta}\]\[=\frac{-tan(ln(\theta))}{\theta}\]Let me know if we need to go over anything.

  9. ammu123
    • 2 years ago
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    thank you this explains it :)

  10. stamp
    • 2 years ago
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    Ok good luck in calculus

  11. ammu123
    • 2 years ago
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    thank you its really hard. just trying to keep up. have a test coming up i just hope to pass it. thank you for ur help!

  12. stamp
    • 2 years ago
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    Sure thing. Private message me if you ever want to discuss more problems.

  13. ammu123
    • 2 years ago
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    ok thank you!

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