Question

y = √1+e power x/ 1-e power x​

Answer 1

Answer:

$\displaystyle y=\sqrt{\frac{1+e^x}{1-e^x}}\\\\\Rightarrow (1-e^x)y^2=1+e^x\\\\\Rightarrow (1-e^x)2y\,dy-e^xy^2\,dx= e^x\,dx\\\\\Rightarrow (1-e^x)^2\,2y\,dy-e^x(1-e^x)y^2\,dx=e^x(1-e^x)\,dx\\\\\Rightarrow (1-e^x)^2\,2y\,dy-e^x(1+e^x)\,dx=e^x(1-e^x)\,dx\\\\\Rightarrow (1-e^x)^2\,2y\,dy=2e^x\,dx$

$\displaystyle\Rightarrow\frac{dy}{dx}=\frac{e^x}{(1-e^x)^2}\times\frac1y\\\\=\frac{e^x}{(1-e^x)^2}\sqrt{\frac{1-e^x}{1+e^x}}\\\\=\frac{-1}{(1-e^x)(1-e^{-x})}\sqrt{\frac{1-e^x}{1+e^x}}\\\\=\frac1{(e^{-x}-1)\sqrt{1-e^{2x}}}$