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Edited by Blonic :
3/6/2014 12:33:43 AM
5
Quick Calculus help. (QUESTION HAS BEEN ANSWERED)
Just wondering, could anyone quickly demonstrate how you solve this problem?
Find the derivative of y in respect to theta (I will call it T).
y=ln(T-8)
I do know you split up ln so that y = (lnT)/(ln8)
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3 RepliesCalculus tutor here. The standard form of a natural log derivative is ln(x) -> 1/x. In this case, we have ln(T-8), so we'll need to use the chain rule. Our outer function is ln(inner), our inner function is T-8, the derivative of the outer function is 1/(inner), the derivative of the inner function is just 1, so putting it all together we end up with 1/(T-8).
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Edited by A Kinky Milkman: 3/6/2014 12:31:57 AMthe derivative of a natural logarithm is (1/x) and usage of the chain rule. T-8 gives us 1 when derived, so there is no need to include it in the final answer. The derivative here is (dy/dT) = (1/(T-8))
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I haven't done anything like that in years. thank god for uni. *shrugs*
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Edited by Mourner: 3/6/2014 12:27:10 AM[quote]Just wondering, could anyone quickly demonstrate how you solve this problem? Find the derivative of y in respect to theta (I will call it T). y=ln(T-8)[/quote] Uhh.. put e^ln(T-8)? Lol idk, I know that the e and ln cancel out.
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I don't think it's the best idea to be asking us for help
