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

Calculus1
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its very simple
please help me
-(1/cos(ln theta))*(sin(ln theta)*1/theta

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Other answers:

\[\frac{d(ln(cos(ln(\theta)))}{dx}\]
Should be dTheta
the answer is - (tan (ln theta))/ theta
i just dont no how to get to that
\[\frac{1}{cos(ln(\theta))}*-sin(ln(\theta))*\frac{1}{\theta}\]\[\frac{-sin(ln(\theta))}{cos(ln(\theta))}\frac{1}{\theta}\]\[=\frac{-tan(ln(\theta))}{\theta}\]Let me know if we need to go over anything.
thank you this explains it :)
Ok good luck in calculus
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!
Sure thing. Private message me if you ever want to discuss more problems.
ok thank you!

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