A) \[{{(-\sin x+{{e}^{x}})}^{-1}}\]
B) \[\frac{\sin x-{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{2}}}\]
C) \[\frac{\sin x-{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{3}}}\]
D) \[\frac{\sin x+{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{3}}}\]
Correct Answer: C
Solution :
\[y=\sin x+{{e}^{x}}\]Þ \[\frac{dy}{dx}=\cos x+{{e}^{x}}\] Þ \[\frac{dx}{dy}={{(\cos x+{{e}^{x}})}^{-1}}\] ?..(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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