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

Question

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

TuteeHUB · Accepted Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

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

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