Respostas

2013-09-29T04:38:37-03:00
f(x) =  \frac{e^x + e^{-x}}{e^x - e^{-x}}

Primeiramente devemos saber as regras da derivada de 

f(x) = e^x

f_{'}(x) = e^x

f(x) = e^{-x}

f_{'}(x) = -e^{-x}

f(x) = \frac{g(x)}{h(x)}

f_{'}(x) = \frac{g_{'}(x) . h(x) - g(x) . h_{'}(x)}{h^2(x)}

Assim:

f(x) = \frac{e^x + e^{-x}}{e^x - e^{-x}}

f_{'}(x) = \frac{(e^x - e^{-x})(e^x - e^{-x}) - (e^x + e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2}

f_{'}(x) = \frac{(e^x)^2 - 2 . (e^x)(e^{-x}) + (e^{-x})^2 - (e^x)^2 - 2(e^x)(e^{-x}) - (e^{-x})^2}{(e^x)^2 - 2(e^x)(e^{-x}) + (e^{-x})^2}

f_{'}(x) = \frac{-2 - 2}{e^{2x} - 2 + \frac{1}{e^{2x}}}

f_{'}(x) = \frac{-4}{\frac{e^{2x} . e^{2x} - 2e^{2x} + 1}{e^{2x}}}

f_{'}(x) = \frac{-4}{\frac{e^{4x} - 2e^{2x} + 1}{e^{2x}}}

f_{'}(x) = \frac{-4e^{2x}}{e^{4x} - 2e^{2x} + 1}

f_{'}(x) = \frac{-4e^{2x}}{(e^{2x} - 1)^2}

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