Step-by-step explanation:
Given that,
[tex]y=\dfrac{e^x+1}{1-e^x}[/tex]
We need to find dy/dx
[tex]\dfrac{dy}{dx}=\dfrac{d}{dx}(\dfrac{e^x+1}{1-e^x})\\\\=\dfrac{1-e^x\dfrac{d}{dx}(e^x+1)-(e^x+1)\dfrac{d}{dx}(1-e^x))}{(1-e^x)^2}\\\\\because \dfrac{d(e^x)}{dx}=e^x\\\\=\dfrac{1-e^x(e^x)-(e^x+1)(-e^x)}{(1-e^x)^2}\\\\=\dfrac{e^x(e^x+1)+(1-e^x)e^x}{(1-e^x)^2}\\\\=\dfrac{\mathrm{e}^x\left(\mathrm{e}^x+1\right)+\left(1-\mathrm{e}^x\right)\mathrm{e}^x}{\left(1-\mathrm{e}^x\right)^2}\\\\=\dfrac{2\mathrm{e}^x}{\left(\mathrm{e}^x-1\right)^2}[/tex]
Hence, this is the required solution.
