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ammu123
Group Title
find the derivative of y with respect to x, t or theta as appropriate.
y= ln(cos(ln theta))
 one year ago
 one year ago
ammu123 Group Title
find the derivative of y with respect to x, t or theta as appropriate. y= ln(cos(ln theta))
 one year ago
 one year ago

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fanitariq Group TitleBest ResponseYou've already chosen the best response.0
its very simple
 one year ago

ammu123 Group TitleBest ResponseYou've already chosen the best response.1
please help me
 one year ago

fanitariq Group TitleBest ResponseYou've already chosen the best response.0
(1/cos(ln theta))*(sin(ln theta)*1/theta
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{d(ln(cos(ln(\theta)))}{dx}\]
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
Should be dTheta
 one year ago

ammu123 Group TitleBest ResponseYou've already chosen the best response.1
the answer is  (tan (ln theta))/ theta
 one year ago

ammu123 Group TitleBest ResponseYou've already chosen the best response.1
i just dont no how to get to that
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{cos(ln(\theta))}*sin(ln(\theta))*\frac{1}{\theta}\]\[\frac{sin(ln(\theta))}{cos(ln(\theta))}\frac{1}{\theta}\]\[=\frac{tan(ln(\theta))}{\theta}\]Let me know if we need to go over anything.
 one year ago

ammu123 Group TitleBest ResponseYou've already chosen the best response.1
thank you this explains it :)
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
Ok good luck in calculus
 one year ago

ammu123 Group TitleBest ResponseYou've already chosen the best response.1
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!
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.1
Sure thing. Private message me if you ever want to discuss more problems.
 one year ago

ammu123 Group TitleBest ResponseYou've already chosen the best response.1
ok thank you!
 one year ago
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