If y = 1 + e^x 1 - e^x , then dy dx = - Vidyadip

MCQ

If $y = \sqrt {{{1 + {e^x}} \over {1 - {e^x}}}} $, then ${{dy} \over {dx}} = $

✓
${{{e^x}} \over {(1 - {e^x})\sqrt {1 - {e^{2x}}} }}$
B
${{{e^x}} \over {(1 - {e^x})\sqrt {1 - {e^x}} }}$
C
${{{e^x}} \over {(1 -{e^x})\sqrt {1 + {e^{2x}}} }}$
D
${{{e^x}} \over {(1 + {e^x})\sqrt {1 - {e^{2x}}} }}$

✓

Answer

Correct option: A.

${{{e^x}} \over {(1 - {e^x})\sqrt {1 - {e^{2x}}} }}$

a
(a) $y = \sqrt {\frac{{1 + {e^x}}}{{1 - {e^x}}}} $ or ${y^2} = \frac{{1 + {e^x}}}{{1 - {e^x}}}$

$2y\frac{{dy}}{{dx}} = \frac{{(1 - {e^x}){e^x} + (1 + {e^x}){e^x}}}{{{{(1 - {e^x})}^2}}} = \frac{{2{e^x}}}{{{{(1 - {e^x})}^2}}}$

$\therefore \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{{(1 - {e^x})}^2}}}\sqrt {\left[ {\frac{{1 - {e^x}}}{{1 + {e^x}}}} \right]\left[ {\frac{{1 - {e^x}}}{{1 - {e^x}}}} \right]} $

$ = \frac{{{e^x}}}{{(1 - {e^x})\sqrt {1 - {e^{2x}}} }}$.

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