Question

Calcule a derivada da função y= ln |ax+b|

Answer 1

Explicação passo-a-passo:

$\sf y=ln~|ax+b|$

$\sf y=\begin{cases} \sf ln~(ax+b),~se~x\ge\dfrac{-b}{a} \\ \\ \sf ln~(-ax-b),~se~x < \dfrac{-b}{a}\end{cases}$

$\sf y'=\begin{cases} \sf \dfrac{(ax+b)'}{ax+b},~se~x\ge\dfrac{-b}{a} \\ \\ \sf \dfrac{(-ax-b)'}{-ax-b},~se~x< \dfrac{-b}{a}\end{cases}$

$\sf y'=\begin{cases} \sf \dfrac{a}{ax+b},~se~x\ge\dfrac{-b}{a} \\ \\ \sf \dfrac{-a}{-ax-b},~se~x < \dfrac{-b}{a}\end{cases}$

$\sf y'=\begin{cases} \sf \dfrac{a}{ax+b},~se~x\ge\dfrac{-b}{a} \\ \\ \sf \dfrac{a}{ax+b},~se~x < \dfrac{-b}{a}\end{cases}$

$\sf y'=\dfrac{a}{ax+b}$