3x/(x+8).
I need to use the 2nd derivative test to find the point(s) of inflection. I get confused about how to do derivatives of examples like this.
It is OK for f'(x)
BTW an easiest way in this case is $\displaystyle f(x) = \frac{3x}{x+8} = 3 - \frac{24}{x+8}$. The derivative of f is the same as the derivative of $\displaystyle -\frac{24}{x+8}$ which is $\displaystyle \frac{24}{(x+8)^2}$
The derivative of $\displaystyle u^n(x)$ is $\displaystyle n u'(x) u^{n-1}(x)$
Use this formula with u(x) = x+8 and n=-2 to get the derivative of $\displaystyle \frac{1}{(x+8)^2}$ then multiply by 24
$\displaystyle f'(x)= \frac{24}{x^2 + 16x + 64}$ then
$\displaystyle f''(x)= \frac{(x^2+16x+64) \cdot 0 - 24(2x+16)}{ (x^2 + 16x + 64)^2} = \boxed{\frac{ - 48x-384}{ x^4 + 32·x^3 + 384·x^2 + 2048·x + 4096} }$
This result is correct but nearly useless. Compare with the method posted by running-gag:
$\displaystyle f'(x)= \frac{24}{x^2 + 16x + 64} = 24 (x+8)^{-2}$ then
$\displaystyle f'(x)= -48 (x+8)^{-3} \cdot 1 =\boxed{ -\frac{48}{(x+8)^3}}$
Both terms describe the same function!