Exercise [06.09]

chag-art · 13 May 2008, 08:58

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 :

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

Exercise [06.09]

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