y=ex−e−xex+e−x then dydx=

dydx=e−x(e2x−1)e−x(e2x+1) =(e2x+1)2e2x−(e2x−1)2e2x(e2x+1)2 =2e2x(e2x+1−e2x+1)(e2x+1)2 =4e2x(e2x+1)2

y=ex−e−xex+e−x then dydx=

dydx=e−x(e2x−1)e−x(e2x+1) =(e2x+1)2e2x−(e2x−1)2e2x(e2x+1)2 =2e2x(e2x+1−e2x+1)(e2x+1)2 =4e2x(e2x+1)2

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$y = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} t h e n \frac{d y}{d x} =$

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$\frac{4 e^{2 x}}{{(e^{2 x} + 1)}^{2}}$

$\frac{e^{2 x}}{e^{2 x} + 1}$

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

$\frac{e^{x}}{{(e^{x} + 1)}^{2}}$

answer is B.

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Detailed Solution

$\frac{d y}{d x} = \frac{e^{- x} (e^{2 x} - 1)}{e^{- x} (e^{2 x} + 1)} = \frac{(e^{2 x} + 1) 2 e^{2 x} - (e^{2 x} - 1) 2 e^{2 x}}{{(e^{2 x} + 1)}^{2}} = \frac{2 e^{2 x} (e^{2 x} + 1 - e^{2 x} + 1)}{{(e^{2 x} + 1)}^{2}} = \frac{4 e^{2 x}}{{(e^{2 x} + 1)}^{2}}$