Question

Alguém me ajuda a achar a derivada dessas duas questões?
a) y = e ^ x . sen (-x)

b) y = ln (x-1 / x²)

Answer 1

Vamos utilizar regra da cadeia e propriedades para funções logarítmicas:

I -

$y'=e^xsen(-x) \\ \\ y=-e^xsen(x)\\ \\ y=-e^x'sen(x)-e^xsen(x)' \\ \\ y=-e^xsen(x)-e^xcos(x).(x)' \\ \\ y=-e^xsen(x)-e^xcos(x).(1) \\ \\ y=-e^xsen(x)-e^xcos(x) \\ \\ y=-e^xcos(x)-e^xsen(x)$

II -

$y=ln(\frac{x-1}{x^2}) \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.(\frac{x-1}{x^2})' \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.\frac{(x-1)'x^2-(x-1)x^2'}{(x^2)^2} \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.\frac{(1-0).x^2-(x-1)2x}{x^4} \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.\frac{x^2-(x-1)2x}{x^4} \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.\frac{x^2-(2x^2-2x)}{x^4} \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.\frac{x^2-2x^2+2x}{x^4} \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.\frac{-x^2+2x}{x^4} \\ \\ y=\frac{1}{\frac{x-1}{x^2}}.\frac{x(-x+2)}{x(x^{3})}$

$y=\frac{1}{\frac{x-1}{x^2}}.\frac{-x+2}{x^{3}} \\ \\ y=\frac{-x+2}{x^{3}(\frac{x-1}{x^2})} \\ \\ y=\frac{-x+2}{\frac{x^4-x^3}{x^2}} \\ \\ y=\frac{-x+2}{\frac{x^2(x^2-x)}{x^2}} \\ \\ y=\frac{-x+2}{x^2-x} \\ \\ y=\frac{-x+2}{x(x-1)}$