Continuity and Differentiability — Maths STD 12 Science — Question

Taking log on both sides, we have

$ \log y = \frac{1}{2}\left[ {\log \left( {x - 1} \right) + \log \left( {x - 2} \right) - \log \left( {x - 3} \right) - \log \left( {x - 4} \right) - \log \left( {x - 5} \right)} \right]$

Differentiating both sides w.r.t x, we get
$\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{1}{{x - 1}}\frac{d}{{dx}}\left( {x - 1} \right) + \frac{1}{{x - 2}}\frac{d}{{dx}}\left( {x - 2} \right) - \frac{1}{{x - 3}}\frac{d}{{dx}}\left( {x - 3} \right) - \frac{1}{{x - 4}}\frac{d}{{dx}}\left( {x - 4} \right) - \frac{1}{{x - 5}}\frac{d}{{dx}}\left( {x - 5} \right)} \right]$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{2}y\left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} -\frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]$

$\Rightarrow \frac{{dy}}{{dx}} = \frac{1}{2}\sqrt {\frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)\left( {x - 5} \right)}}} \left[ {\frac{1}{{x - 1}} + \frac{1}{{x - 2}} - \frac{1}{{x - 3}} - \frac{1}{{x - 4}} - \frac{1}{{x - 5}}} \right]$