$\cfrac{{{{\cos }^{ - 1}}\left( {\cfrac{x}{2}} \right)}}{{\sqrt {2x + 7} }}, - 2 < x < 2.$

Let $y = {\cos ^{ - 1}}\cfrac{x}{2}{(2x + 7)^{ - 1/2}}$ ..(i)

Differentiating (i) w.r.t. x, we get

$\cfrac{{dy}}{{dx}} = {\cos ^{ - 1}}\cfrac{x}{2}\left[ {\cfrac{d}{{dx}}{{(2x + 7)}^{ - 1/2}}} \right] + {(2x + 7)^{ - 1/2}}\left( {\cfrac{d}{{dx}}{{\cos }^{ - 1}}\cfrac{x}{2}} \right)$

$= {\cos ^{ - 1}}\cfrac{x}{2}\left[ {\cfrac{{ - 1}}{2}{{(2x + 7)}^{ - 3/2}}(2)} \right]$ $+ {(2x + 7)^{ - 1/2}}\left( {\cfrac{{ - 1}}{{\sqrt {1 - {{\left( {\cfrac{x}{2}} \right)}^2}} }}} \right) \times \cfrac{1}{2}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = - {\cos ^{ - 1}}\cfrac{x}{2}{(2x + 7)^{ - 3/2}} + {(2x + 7)^{ - 1/2}}\left[ {\cfrac{{ - 1}}{{2\sqrt {1 - \left( {\cfrac{{{x^2}}}{4}} \right)} }}} \right]$

$= - \left[ {\cfrac{1}{{\sqrt {4 - {x^2}} \sqrt {2x + 7} }} + \cfrac{{{{\cos }^{ - 1}}\cfrac{x}{2}}}{{{{(2x + 7)}^{3/2}}}}} \right]$