If \[y=\frac{{{e}^{2x}}+{{e}^{-2x}}}{{{e}^{2x}}-{{e}^{-2x}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1988]

\[\frac{8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

\[\frac{-8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

\[\frac{-4}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

\[\frac{4}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

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

1.	If \[y=\frac{{{e}^{2x}}+{{e}^{-2x}}}{{{e}^{2x}}-{{e}^{-2x}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1988]
A.	\[\frac{-8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]
B.	\[\frac{8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]
C.	\[\frac{-4}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]
D.	\[\frac{4}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]
Answer» B. \[\frac{8}{{{({{e}^{2x}}-{{e}^{-2x}})}^{2}}}\]

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If \[y=\frac{{{e}^{2x}}+{{e}^{-2x}}}{{{e}^{2x}}-{{e}^{-2x}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1988]

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