Find the derivative of $-\frac{x}{2}log_2(\frac{x}{2}) - \left(1 -\frac{x}{2}\right)log_2\left(1 -\frac{x}{2}\right)$.

Question

I get this solving a problem of information theory, this is of the entropy of $H(\frac{x}{2})$.

I know the result is $$\frac12 \log_2 \left(\frac {1-x/2}{x/2}\right)~.$$

But I don't get it.

I don't know what I'm doing wrong but I have this:

$$ - \left[ \frac{1}{2} \left( \log_2(x)-1\right) + \frac{x}{2}\left(\frac{1}{ \ln(2)x}\right) -\frac{1}{2} \bigl( \log_2(2-x)-1 \bigr) + \bigl( \frac{x}2 - 1 \bigr)\left(\frac{1}{ \ln(2)(2-x)}-1 \right) \right]$$

Answer 1

It is worth considering the differentiation of the general case $$\frac{d}{dz}\left[z \log_2 z\right] = \left(1 \cdot \log_2 z \right) + \left( z \cdot \frac{1}{z} \cdot \frac{1}{\log 2} \right) = \log_2 z + \frac{1}{\log 2}.$$ Consequently, with $z = x/2$, we have $$\frac{d}{dx} \left[\frac{x}{2} \log_2 \frac{x}{2}\right] = \left(\log_2 \frac{x}{2} + \frac{1}{\log 2}\right) \frac{d}{dx}\left[\frac{x}{2}\right] = \frac{1}{2} \left( \log_2 \frac{x}{2} + \frac{1}{\log 2} \right),$$ and similarly, with $z = 1 - \frac{x}{2}$, $$\frac{d}{dx}\left[\left(1 - \frac{x}{2}\right) \log_2 \left(1 - \frac{x}{2}\right) \right] = -\frac{1}{2}\left(\log_2 \left( 1 - \frac{x}{2} \right) + \frac{1}{\log 2} \right).$$ It follows that $$\frac{d}{dx} \left[-\frac{x}{2}\log_2 \frac{x}{2} - \left(1 - \frac{x}{2}\right)\log_2 \left(1 - \frac{x}{2}\right) \right] = -\frac{1}{2} \log_2 \frac{x}{2} + \frac{1}{2} \log_2 \left(1 - \frac{x}{2}\right). $$ Note the constants cancel out. This is further simplified to $$\frac{1}{2} \log_2 \frac{1 - x/2}{x/2}$$ as claimed.

Answer 2

$$ \text{We need to find } {d \over {dx}}\left[ { - {x \over 2}\log _2 \left( {{x \over 2}} \right) - \left( {1 - {x \over 2}} \right)\log _2 \left( {1 - {x \over 2}} \right)} \right] \text{.} $$ $$ {\text{Let }}u = {x \over 2} \text{ (Just to abbreviate, as this is a lot of work!) }.{\text{ Then }}u'\left( x \right) = u' = {1 \over 2}.{\text{ So:}} $$ $$ f'\left( x \right) = - {d \over {dx}}\left[ {u\log _2 \left( u \right) + \left( {1 - u} \right)\log _2 \left( {1 - u} \right)} \right] $$ $$ = - {d \over {dx}}\left[ {u\log _2 \left( u \right) - u\log _2 \left( {1 - u} \right) + \log _2 \left( {1 - u} \right)} \right] $$ $$ = - {d \over {dx}}\left[ {u\left( {\log _2 \left( u \right) - \log _2 \left( {1 - u} \right)} \right) + \log _2 \left( {1 - u} \right)} \right] $$ $$ = - {d \over {dx}}\left[ {u\log _2 \left( {{u \over {1 - u}}} \right) + \log _2 \left( {1 - u} \right)} \right] $$ $$=- \left\{ {\left( u \right)\left[ {{1 \over {\ln \left( 2 \right)}}{{{{\left( {1 - u} \right)\left[ {u'} \right] - \left[ {0 - u'} \right]\left( u \right)} \over {\left( {1 - u} \right)^2 }}} \over {\left( {{u \over {1 - u}}} \right)}}} \right] + \left[ {u'} \right]\left( {\log _2 \left( {{u \over {1 - u}}} \right)} \right) + {1 \over {\ln \left( 2 \right)}}{{0 - u'} \over {1 - u}}} \right\} $$ $$ = - \left\{ {u\left[ {{1 \over {\ln \left( 2 \right)}}{{\left( {1 - u} \right)\left[ {u'} \right] + \left[ {u'} \right]u} \over {\left( {1 - u} \right)^2 }}\left( {{{1 - u} \over u}} \right)} \right] + u'\log _2 \left( {{u \over {1 - u}}} \right) - {1 \over {\ln \left( 2 \right)}}{{u'} \over {1 - u}}} \right\} $$ $$ = - \left\{ {u\left[ {{1 \over {\ln \left( 2 \right)}}{{u'\left( {1 - u + u} \right)} \over {\left( {1 - u} \right)^2 }}\left( {{{1 - u} \over u}} \right)} \right] + u'\log _2 \left( {{u \over {1 - u}}} \right) - {1 \over {\ln \left( 2 \right)}}{{u'} \over {1 - u}}} \right\} $$ $$ = - \left\{ {u\left[ {{1 \over {\ln \left( 2 \right)}}{{u'} \over {\left( {1 - u} \right)^2 }}\left( {{{1 - u} \over u}} \right)} \right] + u'\log _2 \left( {{u \over {1 - u}}} \right) - {1 \over {\ln \left( 2 \right)}}{{u'} \over {1 - u}}} \right\} $$ $$ = - \left\{ {{1 \over {\ln \left( 2 \right)}}{{u'} \over {\left( {1 - u} \right)}} + u'\log _2 \left( {{u \over {1 - u}}} \right) - {1 \over {\ln \left( 2 \right)}}{{u'} \over {1 - u}}} \right\} $$ $$ = - u'\left\{ {{1 \over {\ln \left( 2 \right)}}{1 \over {1 - u}} + \log _2 \left( {{u \over {1 - u}}} \right) - {1 \over {\ln \left( 2 \right)}}{1 \over {1 - u}}} \right\} $$ $$ = \left( { - 1} \right)u'\left\{ {\log _2 \left( {{u \over {1 - u}}} \right)} \right\} = \left( { - 1} \right)\left( {{1 \over 2}} \right)\log _2 \left( {{{{x \over 2}} \over {1 - {x \over 2}}}} \right) $$ $$ = {1 \over 2}\left( { - 1} \right)\log _2 \left( {{{{x \over 2}} \over {1 - {x \over 2}}}} \right) = {1 \over 2}\log _2 \left( {\left( {{{{x \over 2}} \over {1 - {x \over 2}}}} \right)^{ - 1} } \right) $$ $$ = {1 \over 2}\log _2 \left( {{{1 - x/2} \over {x/2}}} \right) $$