STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

==> $\frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\sqrt {{a^2} + {x^2}} + x\frac{1}{2}{{({a^2} + {x^2})}^{ - 1/2}}2x} \right]$

$ + \frac{{{a^2}}}{2}\frac{1}{{(x + \sqrt {({x^2} + {a^2})} }}\left[ {1 + \frac{1}{2}{{({x^2} + {a^2})}^{ - 1/2}}2x} \right]$

$ = \frac{{1}}{{ 2 (\sqrt {{a^2} + {x^2}} )}} [ a^2 + 2x^2 + a^2 ]$

$= \frac{{ 2 ( {{a^2} + {x^2}} ) }}{{ 2 (\sqrt {{a^2} + {x^2}} )}} $

$= \sqrt {{a^2} + {x^2}} $