A) \[-x\]
B) \[x\]
C) \[y\]
D) \[-y\]
Correct Answer: C
Solution :
Here we have \[y=1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}....\] \[y={{e}^{-x}}\left[ 1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}....={{e}^{-x}} \right]\] \[\therefore \] \[\frac{dy}{dx}=-{{e}^{-x}}\] \[\therefore \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{-x}}=y\]You need to login to perform this action.
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