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Find the derivative of the function y=ln (1+e^x/ 1-e^x)
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can we use logarithmic properties here?
chain rule
yes, but first we can break it up and make it easier on ourself
ya.it would be better ....using log both side
2e^x/(1-e^2x)
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yes how did you get that
yeah thats the answer but to get there we can write it as: y=ln(1+e^x)-ln(1-e^x)
Then we differntiate both sides
\[y=\ln(1+e^x)-\ln|1-e^x|=>y'=\frac{(1+e^x)'}{1+e^x}-\frac{(1-e^x)'}{1-e^x}\]
y=sin\[\sqrt[3]{x}\] +\[\ \sqrt[3]{sinx}\]
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