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

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reemii
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0were you able to do it??

reemii
 2 years 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.
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0If that's it yea easy enough

reemii
 2 years 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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