Find the second derivative of y when y is given in terms of x. (Solved) y = $(2+1/x)^3$
Find y''.
Explanation for help:
The correct answer is y'' = $6/x^3(2+1/x)(2+2/x)$.
So far, I'm at y' = $-3/x^2(2+1/x)^2$.
I'm not sure how to get from y' to y'', though.
Could someone please show how to solve for y'' starting from y'?

This problem has been solved.


Starting with $y' = −3/x^2(2+1/x)^2$, we solve for $y''$,
$$
y'' = \frac{\mathrm{d}}{\mathrm{d}x}\Big[-3/x^2\Big](2+1/x)^2 + \frac{\mathrm{d}}{\mathrm{d}x}\Big[(2+1/x)^2\Big](-3/x^2).
$$
We now simplify and then factor.
\begin{array}{rl}
  y'' = & 6/x^3\cdot(2+1/x)^2 + (-2/x^2)\cdot(2+1/x)\cdot(-3/x^2) \\
      = & 6/x^3\cdot(2+1/x)^2 + 6/x^4\cdot(2+1/x) \\
      = & 6/x^3\cdot\Big((2+1/x)^2+1/x\cdot(2+1/x)\Big) \hspace{10pt}\text{($6/x^3$ was factored out here)} \\
      = & 6/x^3\cdot(4+4/x+1/x^2+2/x+1/x^2) \\
      = & 6/x^3\cdot(2/x^2+6/x+4) \\
      = & 12/x^3\cdot(1/x^2+3/x+2) \\
      = & 12/x^3\cdot(1/x+2)\cdot(1/x+1) \\
\end{array}
Hence we have

$$ y'' = 6/x^3\cdot(2+1/x)\cdot(2+2/x). $$

