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.

 A: Hint: The chain rule is
$$
\frac{\mathrm{d}}{\mathrm{d}x}\Big[ f\big(g(x)\big) \Big] = f'\big(g(x)\big)\cdot g'(x),
$$
the product rule is
$$
\frac{\mathrm{d}}{\mathrm{d}x}\Big[ f(x)\cdot g(x) \Big] = f'(x)\cdot g(x) + f(x)\cdot g'(x),
$$
and you can always write $1/x$ as $x^{-1}$.
Note that $f'(x)$ means $\frac{\mathrm{d}}{\mathrm{d}x}\big[f(x)\big]$ and thus $y' = \frac{\mathrm{d}}{\mathrm{d}x}[y]$.

This is not a homework solving site, but I'll start you off.
Write
$$
y = (2 + x^{-1})^{3}.
$$
Apply the chain rule to get $y' = \frac{\mathrm{d}}{\mathrm{d}x}[y]$. In this case, you have $f(x) = x^{3}$ and $g(x) = 2 + x^{-1}$, so
$$
\frac{\mathrm{d}}{\mathrm{d}x}\Big[ (2 + x^{-1})^{3} \Big] = 3\cdot (2 + x^{-1})^{2} \cdot \frac{\mathrm{d}}{\mathrm{d}x}\big[ 2 + x^{-1} \big] = 3\cdot (2 + x^{-1})^{2} \cdot (-x^{-2}).
$$
Thus, we have that
$$
y' = \frac{\mathrm{d}}{\mathrm{d}x}\Big[ (2 + x^{-1})^{3} \Big] = -3x^{-2}\cdot (2 + x^{-1})^{2}.
$$
See if you can do the second derivative by letting $f(x) = -3x^{-2}$ and $g(x) = (2 + x^{-1})^{2}$ using the product rule (and the chain rule again on $g(x)$).
A: Hint: Write $$y^{1/3}=\frac{2x+1}{x}$$ then you will get
$$\frac{1}{3}y^{-2/3}y'=\frac{2x-(2x+1)}{x^2}$$
and
$$\frac{1}{3}(-\frac{2}{3})y^{-5/3}(y')^2+\frac{1}{3}y^{-2/3}y''=2x^{-3}$$
For your control: The second derivative is given by $$f''(x)=12\,{\frac { \left( x+1 \right)  \left( 2\,x+1 \right) }{{x}^{5}}}$$
A: 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

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

