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Christos
Group Title
Logarithmic differantiation,
http://screencast.com/t/fwBW2D9z9E
Can you help me out on this please
 one year ago
 one year ago
Christos Group Title
Logarithmic differantiation, http://screencast.com/t/fwBW2D9z9E Can you help me out on this please
 one year ago
 one year ago

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reemii Group TitleBest ResponseYou've already chosen the best response.0
when there is an \(x\) in the exponent and want to compute a limit or a derivative, always resort to this technique: \[ a^{h(x)} = e^{\ln a^{h(x)}} = e^{h(x) \ln a} \] from here you should be able to do something.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
were you able to do it??
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
do you know the "more simple" formula: \((a^x)' = a^x \ln a\) ? (\(a>0\)) knowing this you don't need the above trick. (you have to know the derivation of a compound function formula)
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.1
Let y=f(x) \[\Large y=\pi^{sinx}\] ln both sides \[\ln(y)=\sin(x)\ln(\pi)\] Differentiate each term \[\frac{1}{y}y'=\ln(\pi)\cos(x) \\ \\ y'=yln(\pi)\cos(x) \\ \\ y'=\pi^{sinx}\ln(\pi)\cos(x)\] Got it?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
If that's it yea easy enough
 one year ago

reemii Group TitleBest ResponseYou've already chosen the best response.0
("always" means that it works, but there can be simpler ways depending on the problem) \[ (\pi^{\sin x})' = (\pi^{\sin x} \ln \pi) (\sin x)' = (\pi^{\sin x} \ln \pi) (\cos x) \] (derivation by parts.)
 one year ago
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