Derivative of Ln (1 + e ^ - 2x) ^ 1 / 2 | Math enthusiast | Mathematical art

Derivative of Ln (1 + e ^ - 2x) ^ 1 / 2

y=ln√(1+e^-2x)
=1/2 ln(1+e^-2x)
y''=1/2 [ln(1+e^-2x)]'
=1/2 * 1/(1+e^-2x) *(1+e^-2x)'
=1/[2(1+e^-2x)] *-2e^-2x
=-2e^-2x / [2(1+e^-2x)]
=-1/(e^2x+1)

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