I was given an excellent answer to a previous question about finding a second derivative of a function. At the end of the answer the writer said there was an alternative method, using the chain rule. I cannot find how to do this.

I am trying to find the second derivative of function $y = \sqrt\frac{6x}{x + 2}$ when x = 4.

I can follow all of the following:

$y = \sqrt\frac{6x}{x + 2} = \sqrt u$

For first derivative:

$\frac{dy}{dx} = \frac{dy}{du}.\frac{du}{dx} = \frac{1}{2 \sqrt u}.\frac{12}{(x + 2)^2} = \frac {6}{(x + 2)^2}\sqrt \frac{x + 2}{6x}$

$= 6(6x)^{-1/2}(x + 2)^{-3/2}$

Now, this is where I come unstuck.

I know I use the formula $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

Let $u = 6(6x)^{-1/2}, v = (x + 2)^{-3/2}$

I calculate $\frac{dv}{dx}$ = $\frac{-3}{2}(x + 2)^{-5/2}, \frac{du}{dx} = -18(6x)^{-3/2}$

Am I going right so far? My workings turn very messy and I cannot obtain the final answer, which should be -1/32.

  • $\begingroup$ First, you don't say what you are taking to be "u" and what "v". $\endgroup$
    – user247327
    Oct 16 '20 at 16:33
  • $\begingroup$ "The answer should be $-\frac{1}{32}$." Is that the derivative at a specific point? Because this function certainly does not have a constant second derivative... $\endgroup$ Oct 16 '20 at 16:49
  • $\begingroup$ (Your calculations of $\frac{dv}{dx}$ and $\frac{du}{dx}$ are correct, given your $u$ and $v$; I did not check if your $u$ and $v$ are correct.) $\endgroup$ Oct 16 '20 at 16:50
  • $\begingroup$ Sorry. When x = 4 $\endgroup$
    – Steblo
    Oct 16 '20 at 16:50
  • $\begingroup$ Then just plug in $x=4$ into $u$, $v$, $u'$, and $v'$, and then do the operation. No need to find the general formula if all you need is a single value. $\endgroup$ Oct 16 '20 at 16:58

If $u=6(6x)^{-1/2}$, then $u' = -\frac{6}{2}(6x)^{-3/2}(6x)' = -18(6x)^{-3/2}$.

If $v=(x+2)^{-3/2}$, then $v' = -\frac{3}{2}(x+2)^{-5/2}(x+2)' = -\frac{3}{2}(x+2)^{-5/2}$.

If all you need to do is figure out the value at $x=4$, then don't find a full formula for the second derivative: just plug in $x=4$ into $u$, $v$, $\frac{du}{dx}$ and $\frac{dv}{dx}$ before plugging into the formula for the second derivative.

There is no point in finding the general formula for the second derivative if all you need is a single value.

At $x=4$, $u(4) = 6(24)^{-1/2} = \frac{6}{\sqrt{24}} = \frac{6}{2\sqrt{6}} = \frac{\sqrt{6}}{2}$.

At $x=4$, $u'(4)=-\frac{18}{(24)^{-3/2}} = -\frac{18}{(2\sqrt{6})^3} = -\frac{3}{8\sqrt{6}}$.

At $x=4$, $v(4) = \frac{1}{6^{-3/2}} = \frac{1}{(\sqrt{6})^3}= \frac{1}{6\sqrt{6}}$.

At $x=4$, $v'(4) = -\frac{3}{2(\sqrt{6})^{5}} = -\frac{3}{72\sqrt{6}} = -\frac{1}{24\sqrt{6}}$.

So $$\begin{align*} u\frac{dv}{dx} + v\frac{du}{dx}\Bigm|_{x=4} &= \frac{\sqrt{6}}{2}\left(-\frac{1}{24\sqrt{6}}\right) + \frac{1}{6\sqrt{6}}\left(-\frac{3}{8\sqrt{6}}\right)\\ &= -\frac{1}{48} - \frac{3}{(36)(8)} = -\frac{1}{48}-\frac{1}{96}\\ &= -\frac{2}{96} - \frac{1}{96} = -\frac{3}{96} = -\frac{1}{32}. \end{align*}$$


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