Question

Questão de derivadas, letra p
como resolver e qual a resposta?

Answer 1

$f(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}\\\\\\f(x)=\frac{e^x+\frac{1}{e^x}}{e^x-\frac{1}{e^x}}\\\\\\f'(x)=\frac{\left(e^x+\frac{0\cdot\,e^x-1\cdot\,e^x}{(e^x)^2}\right)\left(e^x-\frac{1}{e^x}\right)-\left(e^x+\frac{1}{e^x}\right)\left(e^x-(\frac{0\cdot\,e^x-1\cdot\,e^x}{(e^x)^2})\right)}{\left (e^x-\frac{1}{e^x}\right)^2}$

$f'(x)=\frac{\left(e^x-\frac{1}{(e^x)^2}\right)\left(e^x-\frac{1}{e^x}\right)-\left(e^x+\frac{1}{e^x}\right)\left(e^x+\frac{1}{(e^x)^2}\right)}{\left(e^x-\frac{1}{e^x}\right)^2}\\\\\\f'(x)=\frac{(e^x)^2-e^x\cdot\frac{1}{e^x}-\frac{1}{(e^x)^2}\cdot\,e^x+\frac{1}{(e^x)^2}\cdot\frac{1}{e^x}-\left[(e^x)^2+e^x\cdot\frac{1}{(e^x)^2}+\frac{1}{e^x}\cdot\,e^x+\frac{1}{e^x}\cdot\frac{1}{(e^x)^2}\right]}{\left(e^x-\frac{1}{e^x}\right)^2}\\\\\\f'(x)=\frac{(e^x)^2-1-\frac{1}{e^x}+\frac{1}{(e^x)^3}-(e^x)^2-\frac{1}{e^x}-1-\frac{1}{(e^x)^3}}{\left (e^x-\frac{1}{e^x}\right)^2}\\\\\\f'(x)=\frac{-2 - \frac{2}{e^x}}{\left(e^x-\frac{1}{e^x}\right)^2}}$