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 Exercise [06.09] 
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Joined: 11 May 2008, 20:07
Posts: 11
Location: france
Post Exercise [06.09]
Before the solution there is a little math relation we need to know :

z=x^y=e^{log{x^y}}=e^{ylogx}<br /><br />logz=ylogx<br /><br />y= log_xz= \frac{logz}{logx}

Now we can proceed, and everything is clear (little remind : a is constant ).

The first one :

d(log_ax)=d(\frac{logx}{loga})=\frac{1}{xloga}=\frac{logx}{logx}\times\frac{1}{x{loga}}=\frac{log_ax}{xlogx}

The second one :

d(log_xa)=d(\frac{loga}{xlogx})=loga\times{d({(logx)}^{-1})}=-\frac{loga}{x{(logx)}^2}=-\frac{log_xa}{xlogx}

The third one :

dx^x=de^{xlogx}=e^{xlogx}d(xlogx)=e^{xlogx}(1+logx)=(1+logx)x^x


13 May 2008, 08:58
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