A) \[{{e}^{2x}}+{{(-1)}^{n}}{{e}^{-2x}}\]
B) \[{{2}^{n}}({{e}^{2x}}-{{e}^{-2x}})\]
C) \[{{2}^{n}}[{{e}^{2x}}+{{(-1)}^{n}}{{e}^{-2x}}]\]
D) None of these
Correct Answer: C
Solution :
\[\frac{d}{dx}[{{e}^{2x}}+{{e}^{-2x}}]=2{{e}^{2x}}+2{{e}^{-2x}}={{2}^{1}}[{{e}^{2x}}-{{e}^{-2x}}]\] \[\frac{{{d}^{2}}}{d{{x}^{2}}}[{{e}^{2x}}+{{e}^{-2x}}]={{2}^{2}}[{{e}^{2x}}+{{e}^{-2x}}]\] \[\frac{{{d}^{2}}}{d{{x}^{2}}}[{{e}^{2x}}+{{e}^{-2x}}]={{2}^{2}}[{{e}^{2x}}-{{e}^{-2x}}]\] ................................................... ................................................... \[\frac{{{d}^{n}}}{d{{x}^{n}}}[{{e}^{2x}}+{{e}^{-2x}}]={{2}^{n}}[{{e}^{2x}}+{{(-1)}^{n}}{{e}^{-2x}}]\].
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