If \[y=\sqrt{\frac{1+{{e}^{x}}}{1-{{e}^{x}}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1986]

\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{x}}}}\]

\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{2x}}}}\]

\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1+{{e}^{2x}}}}\]

\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{2x}}}}\]

If \[y=\sqrt{\frac{1+{{e}^{x}}}{1-{{e}^{x}}}}\], then \[\frac{dy}{dx}=

1.	If \[y=\sqrt{\frac{1+{{e}^{x}}}{1-{{e}^{x}}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1986]
A.	\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{2x}}}}\]
B.	\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{x}}}}\]
C.	\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1+{{e}^{2x}}}}\]
D.	\[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{2x}}}}\]
Answer» B. \[\frac{{{e}^{x}}}{(1-{{e}^{x}})\sqrt{1-{{e}^{x}}}}\]

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If \[y=\sqrt{\frac{1+{{e}^{x}}}{1-{{e}^{x}}}}\], then \[\frac{dy}{dx}=\] [AI CBSE 1986]

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