If y = x + e^x , then d^2 x d y^2 = - Vidyadip

MCQ

If $y = \sin x + {e^x},$ then ${{{d^2}x} \over {d{y^2}}} = $

A
${( - \sin x + {e^x})^{ - 1}}$
B
${{\sin x - {e^x}} \over {{{(\cos x + {e^x})}^2}}}$
✓
${{\sin x - {e^x}} \over {{{(\cos x + {e^x})}^3}}}$
D
${{\sin x + {e^x}} \over {{{(\cos x + {e^x})}^3}}}$

✓

Answer

Correct option: C.

${{\sin x - {e^x}} \over {{{(\cos x + {e^x})}^3}}}$

c
(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}}}$.

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