How do you find the second derivative of $ln |(x + 1)^2(x - 2)^4 / (2x + 9)|$ ?

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Answer 1 · 2016-11-27T17:24:57

Rewrite using properties of logarithms before differentiating.

We'll call the function $f$ .

$f(x) = ln(((x+1)^2(x-2)^4)/(2x+9)) = 2ln(x+1)+4ln(x-2)-ln(2x+9)$

Use $d/dx(lnu) = 1/u (du)/dx$

$f'(x) = 2/(x+1)+4/(x-2)-2/(2x+9)$

Now use $d/dx(1/u) = -1/u^2 (du)/dx$ to get

$f''(x) = -2/(x+1)^2 - 4/(x-2)^2 + 4/(2x+9)^2$

How do you find the second derivative of ln |(x + 1)^2(x - 2)^4 / (2x + 9)|ln |(x + 1)^2(x - 2)^4 / (2x + 9)|?