What is the second derivative of $f(x)=x/(x^2 + 1) ^2$ ?

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Answer 1 · 2015-12-26T17:08:11

$f''(x)=(12x(x^2-1))/(x^2+1)^4$

Use the quotient rule: for a function $f(x)=(g(x))/(h(x))$ ,

$f'(x)=(g'(x)h(x)-h'(x)g(x))/(h(x))^2$

$g(x)=x$
$h(x)=(x^2+1)^2$

$g'(x)=1$
$h'(x)=4x(x^2+1)$

Thus,

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

Simplify.

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

$f'(x)=(1-3x^2)/(x^2+1)^3color(white)(sss)$ First Derivative

To find the second derivative, use the quotient rule again.

$g(x)=1-3x^2$
$h(x)=(x^2+1)^3$

$g'(x)=-6x$
$h'(x)=6x(x^2+1)^2$

Thus,

$f''(x)=(-6x(x^2+1)^3-6x(x^2+1)^2(1-3x^2))/(x^2+1)^6$

Simplify.

$f''(x)=(-6x(x^2+1)^2(x^2+1+1-3x^2))/(x^2+1)^6$

$f''(x)=(-6x(-2x^2+2))/(x^2+1)^4$

$f''(x)=(12x(x^2-1))/(x^2+1)^4color(white)(sss)$ Second Derivative

The main two pitfalls here are

What is the second derivative of f(x)=x/(x^2 + 1) ^2f(x)=x/(x^2 + 1) ^2?