Correct option: C.${{\sin x - {e^x}} \over {{{(\cos x + {e^x})}^3}}}$
(c) $y = \sin x + {e^x}$==> $\frac{{dy}}{{dx}} = \cos x + {e^x}$
==> $\frac{{dx}}{{dy}} = {(\cos x + {e^x})^{ - 1}}$ …..$(i)$
Again, $\frac{{{d^2}x}}{{d{y^2}}} = - {(\cos x + {e^x})^{ - 2}}( - \sin x + {e^x})\frac{{dx}}{{dy}}$.
Substituting the value of $\frac{{dx}}{{dy}}$ from $(i),$
$\frac{{{d^2}x}}{{d{y^2}}} = \frac{{(\sin x - {e^x})}}{{{{(\cos x + {e^x})}^2}}}\,{(\cos x + {e^x})^{ - 1}}$
$ = \frac{{\sin x - {e^x}}}{{{{(\cos x + {e^x})}^3}}}$.
