Question

Y=((1+e^x)¦(1-e^x ))^1/2, help me to find y' plz



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Answer 1

Is that a division sign?

Answer 2

yeah it is, then everything raised to power 1/2

Answer 3

\[\large y=\left(\cfrac{1+e^x}{1-e^x}\right)^{1/2}\]First logarithmic differentiation, and then implicit differentiation.

Answer 4

\[\large \ln (y) = \cfrac{1}{2} [\ln (1+e^x)-\ln(1-e^x)]\]

Answer 5

wow, then what about the next step

Answer 6

Or the long way..\[\large \cfrac{d}{dx}(y) = \cfrac{d}{dx}\left(\cfrac{1+e^x}{1-e^x}\right)^{1/2}\]

Answer 7

\[\large y' = \cfrac{1}{2}\left(\cfrac{1+e^x}{1-e^x}\right)^{-1/2} \cdot \cfrac{e^x(1-e^x)+e^x(1+e^x)}{(1-e^x)^2}\]\[\large \cfrac{e^x(1-e^x)+e^x(1+e^x)}{(1-e^x)^2} = \cfrac{e^x -e^{2x}+ e^x +e^{2x}}{(1-e^x)^2}= \cfrac{2e^x}{(1-e^x)^2}\]So \[\large y' = \cfrac{1}{2\sqrt{\cfrac{1+e^x}{1-e^x}}} \cdot \cfrac{2e^x}{(1-e^x)^2}\]\[\large y' = \cfrac{e^x}{(1-e^x)^2\sqrt{\cfrac{1+e^x}{1-e^x}}}\]

Answer 8

thanks alot, am done