If \[y=\sin x+{{e}^{x}},\]then \[\frac{{{d}^{2}}x}{d{{y}^{2}}}=\] [Karnataka CET 1999; UPSEAT 2001; Kurukshetra CEE 2002]

\[\frac{\sin x+{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{3}}}\]

\[{{(-\sin x+{{e}^{x}})}^{-1}}\]

\[\frac{\sin x-{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{2}}}\]

\[\frac{\sin x-{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{3}}}\]

If \[y=\sin x+{{e}^{x}},\]then \[\frac{{{d}^{2}}x}{d{{y}^{2}}}=\]

1.	If \[y=\sin x+{{e}^{x}},\]then \[\frac{{{d}^{2}}x}{d{{y}^{2}}}=\] [Karnataka CET 1999; UPSEAT 2001; Kurukshetra CEE 2002]
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.	\[\frac{\sin x+{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{3}}}\]
Answer» D. \[\frac{\sin x+{{e}^{x}}}{{{(\cos x+{{e}^{x}})}^{3}}}\]

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If \[y=\sin x+{{e}^{x}},\]then \[\frac{{{d}^{2}}x}{d{{y}^{2}}}=\] [Karnataka CET 1999; UPSEAT 2001; Kurukshetra CEE 2002]

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