0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.