Question

find f(x) and f(x).
f(x) = \frac{x}{x^{2}-6}
f(x) =
f(x) =

Explanation:

Step1: Apply quotient - rule for first derivative

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x$, $u'=1$, $v=x^{2}-6$, and $v' = 2x$. So, $f'(x)=\frac{1\cdot(x^{2}-6)-x\cdot(2x)}{(x^{2}-6)^{2}}=\frac{x^{2}-6 - 2x^{2}}{(x^{2}-6)^{2}}=\frac{-x^{2}-6}{(x^{2}-6)^{2}}$.

Step2: Apply quotient - rule for second derivative

Now, for $y = f'(x)=\frac{-x^{2}-6}{(x^{2}-6)^{2}}$, let $u=-x^{2}-6$, $u'=-2x$, $v=(x^{2}-6)^{2}$, and $v' = 2(x^{2}-6)\cdot2x=4x(x^{2}-6)$. Then $f''(x)=\frac{-2x\cdot(x^{2}-6)^{2}-(-x^{2}-6)\cdot4x(x^{2}-6)}{(x^{2}-6)^{4}}$. Factor out $2x(x^{2}-6)$ from the numerator: $f''(x)=\frac{2x(x^{2}-6)[-(x^{2}-6)+2(x^{2}+6)]}{(x^{2}-6)^{4}}=\frac{2x(x^{2}-6)(-x^{2}+6 + 2x^{2}+12)}{(x^{2}-6)^{4}}=\frac{2x(x^{2}+18)}{(x^{2}-6)^{3}}$.

Answer:

$f'(x)=\frac{-x^{2}-6}{(x^{2}-6)^{2}}$
$f''(x)=\frac{2x(x^{2}+18)}{(x^{2}-6)^{3}}$