Answer:
The derivative of the function :[tex]f(x) = \ln \frac{1 + e^x}{1 - e^x}[/tex]
[tex]f(x)'=\frac{2e^x}{1-e^{2x}}[/tex]
Step-by-step explanation:
[tex]f(x) = \ln \frac{1 + e^x}{1 - e^x}[/tex]
Identity = [tex]\frac{d(\ln x)}{dx}=\frac{1}{x}[/tex]
Quotient rule = [tex]\frac{d(\frac{u}{v})}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]
[tex]f(x)'=\frac{1}{\frac{1 + e^x}{1 - e^x}}\times \frac{(1 - e^x)e^x-(1 + e^x)(-e^x)}{(1 - e^x)^2}\times 1[/tex]
[tex]=\frac{1}{\frac{1 + e^x}{1 - e^x}}\times \frac{e^x(1-e^x+1+e^x)}{(1-e^x)^2}[/tex]
[tex]=\frac{1}{\frac{1 + e^x}{1 - e^x}}\times \frac{e^x(2)}{(1 - e^x)^2}[/tex]
[tex]=\frac{1}{(1 + e^x)(1 - e^x)}\times 2e^x[/tex]
Identity =[tex](a+b)(a-b)=a^2-b^2[/tex]
[tex]=\frac{2e^x}{1-e^{2x}}[/tex]
