Question

calcular derivada de f(x) = ln (1/x + 1/x^2), aplicando regra da cadeia.

Answer 1

Olá

$f(x)=ln( \frac{1}{x} + \frac{1}{x^2} ) \\ \\ \frac{df}{dx} = \frac{df}{du} . \frac{du}{dx} \\ \\ ( \frac{1}{x} + \frac{1}{x^2} )=u \\ \\ \frac{d}{du}(ln(u))~ \cdot \frac{d}{dx} (\frac{1}{x}+ \frac{1}{x^2} ) \\ \\ \frac{d}{du}~ ln(u) = \frac{1}{u} \\ \\ \frac{d}{dx} ( \frac{1}{x} + \frac{1}{x^2} ) ~~~~~ \\ Aplicando~a~regra~da~diferenca \\ \\ \frac{d}{dx}( \frac{1}{x})=- \frac{1}{x^2} ~~~~~~~~+~~~~~~~~~ \frac{d}{dx}( \frac{1}{x^2}) = - \frac{2}{x^3}$



$\frac{1}{u}( -\frac{1}{x^2} - \frac{2}{x^3} ) \\ \\ \\ Substituindo~na~equacao \\ \\ \frac{1}{ \frac{1}{x^2} + \frac{2}{x^3} }(- \frac{1}{x^2}- \frac{2}{x^3} ) \\ \\ \\ \boxed{\boxed{f'(x)=\frac{-x-2}{x(x+1)}}}$