A) e
B) 4e
C) 2e
D) 3e
E) 5e
Correct Answer: B
Solution :
Given, \[y={{e}^{x}}.{{e}^{{{x}^{2}}}}.{{e}^{{{x}^{3}}}}.....{{e}^{{{x}^{n}}}}.....\] \[\Rightarrow \] \[y={{e}^{(x+{{x}^{2}}+....+\infty )}}\] \[\Rightarrow \] \[y={{e}^{\frac{x}{1-x}}}\] \[\Rightarrow \] \[\frac{dy}{dx}={{e}^{\frac{x}{1-x}}}\left[ \frac{(1-x)1-x(-1)}{{{(1-x)}^{2}}} \right]\] \[={{e}^{\frac{x}{1-x}}}.\frac{1}{{{(1-x)}^{2}}}\] At \[x=\frac{1}{2},\] \[{{\left( \frac{dy}{dx} \right)}_{x=\frac{1}{2}}}={{e}^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}}.\frac{1}{{{\left( 1-\frac{1}{2} \right)}^{2}}}\] \[=4e\]You need to login to perform this action.
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