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

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reemii
 one year ago
Best ResponseYou've already chosen the best response.0when 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.

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

reemii
 one year ago
Best ResponseYou've already chosen the best response.0do 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)

.Sam.
 one year ago
Best ResponseYou've already chosen the best response.1Let 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?

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

reemii
 one year ago
Best 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.)
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