What is the second derivative of $f(x)= ln (x^2+2x)$ ?

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Answer 1 · 2016-01-13T13:27:42

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

Use the substitution $t = x^2+2x$
Then $f(x) = lnt$

$(df)/dt = 1/t$
$(dt)/dx =2x +2$

$(df)/dx = (df)/dt * (dt)/dx = (1/(x^2+2x))* (2x+2)$
$f'(x) = (2(x+1))/(x^2+2x)$

We can now use the quotient rule to find the second derivative

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

$f''(x) = (2x^2 +4x -2x^2 - 8x -4)/(x^2+2)^2$
$f''(x) =-4(x+1)/(x^2+2)^2$

What is the second derivative of f(x)= ln (x^2+2x)f(x)= ln (x^2+2x)?