How do you find the second derivative of $y=ln((2x)/(x+3))$ ?

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Answer 1 · 2016-11-18T06:24:30

$1/(x+3)^2-1/x^2$

$y= ln 2 + ln x - ln (x|3)$

$y'=0+1/x-1/(x+3)=x^(-1)-(x+2)^(-1)$

$y''=(-1)x^(-1-1)-(-1)(x+2)^(-1-1)(x+3)'$

$=-1/x^2+1/(x+3)^2(1)$

How do you find the second derivative of y=ln((2x)/(x+3))y=ln((2x)/(x+3)) ?