Question

dada a função y=(x+1)/x2+1) calcule dy/dx​

Answer 1

$\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{\frac{df}{dx}.g(x)-f.\frac{dg}{dx}}{[g(x)]^2}\\\\\frac{d}{dx}[\frac{x+1}{x^2+1}]=\frac{(x+1)'(x^2+1)-(x+1)(x^2+1)'}{(x^2+1)^2}\\\\\frac{d}{dx}[\frac{x+1}{x^2+1}]=\frac{(x^2+1)-(x+1)2x}{(x^2+1)^2}\\\\\frac{d}{dx}[\frac{x+1}{x^2+1}]=\frac{x^2+1-2x^2-2x}{(x^2+1)^2}\\\\\frac{d}{dx}[\frac{x+1}{x^2+1}]=\frac{-x^2-2x+1}{(x^2+1)^2}$

Answer 2

Resposta:

Explicação passo-a-passo:

Sendo y = u/v

$y'=\frac{vu'-uv'}{v^2} \\\\y'=\frac{(x^2+1).(x+1)'-(x+1)(x^2+1)'}{(x^2^+1)^2\\}\\y'=\frac{(x^2+1).1-(x+1)2x}{(x^2+1)^2} \\\\y'=\frac{x^2+1-2x^2-2x}{(x^2+1)^2}\\\\y'=\frac{-2x^2+x^2-2x+1}{(x^2+1)^2}\\\\y'=\frac{-x^2-2x+1}{(x^2+1)^2}$