This is a question I've had a lot of trouble with. I HAVE solved it, however, with a lot of trouble and with an extremely ugly calculation. So I want to ask you guys (who are probably more 'mathematically-minded' so to say) how you would solve this. Keep in mind that you shouldn't use too advanced stuff, no differential equations or similair things learned in college:

Given are the functions $f_p(x) = \dfrac{9\sqrt{x^2+p}}{x^2+2}$. The line $k$ with a slope of 2,5 touches $f_p$ in $A$ with $x_A = -1$. Get the function of k algebraically.

  • I might have used wrong terminology, because English is not my native language, I will hopefully clear up doubts on what this problem is by showing what I did.

First off, I got $[f_p(x)]'$. This was EXTREMELY troublesome, and is the main reason why I found this problem challenging, because of all the steps. Can you guys show me the easiest and especially quickest way to get this derivative?

After that, I filled in $-1$ in the derivative and made the derivative equal to $2\dfrac{1}{2}$, this was also troublesome for me, I kept getting wrong answers for a while, again: Can you guys show me the easiest and especially quickest way to solve this?

After you get p it is pretty straightforward. I know this might sound like a weird question, but it basically boils down to: I need quicker and easier ways to do this. I don't want to make careless mistakes, but because the length of these types of question, it ALWAYS happens. Any tips or tricks regarding this topic in general would be much appreciated too.

Update: A bounty will go to the person with the most clear and concise way of solving this question!

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    $\begingroup$ One thing to simplify taking the derivative is to note that $f_p(x)$ depends only on $x^2$, that is, $f_p(x)=g(x^2)$ with $g(y)=9\sqrt{y+p}/(y+2)$; that makes it easier to get $f_p'(x)=2xg'(x^2)$. $\endgroup$ – joriki Oct 25 '12 at 23:07
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    $\begingroup$ A cool trick is to square the relation: $$f_p^2(x) (x^2 + 2)^2 = 81(x^2 + p) \quad\Longrightarrow$$ $$ 2 f_p(x) f_p'(x)(x^2 + 2)^2 + 4x f_p^2(x^2 +2) = 162x \quad\Longrightarrow$$ \begin{align}f_p'(x) &= \frac{162x - 4x f_p^2(x)(x^2+2)}{2f_p(x)(x^2+2)^2}\\ \\ &= \frac{162x - 324x\frac{x^2 + p}{x^2 +2}}{18(x^2 +2)\sqrt{x^2+p}}\\ \\ &= \frac{9 x (x^2+2) - 18x(x^2+p)}{(x^2 + 2)^2 \sqrt{x^2 + p}}\\ \\ &= \frac{9x(2-2p-x^2)}{(x^2 + 2)^2\sqrt{x^2 + p}}\end{align} $\endgroup$ – Pragabhava Oct 25 '12 at 23:34
  • $\begingroup$ @Pragabhava: That's nice, but I daresay not simpler than doing it directly :-) $\endgroup$ – joriki Oct 26 '12 at 1:34
  • $\begingroup$ @joriki True, but maybe if we join forces? Radicals are always a problem for me :) $\endgroup$ – Pragabhava Oct 26 '12 at 1:40
  • $\begingroup$ I would take the derivative of the $\ln$ of $f_p$ and then use that $\ln' f_p(x) = f'_p(x)/f_p(x)$. However, I'm not sure if logs are already considered too advanced stuff? $\endgroup$ – Fabian Oct 27 '12 at 23:11

By "touches" I assume you mean that the line is tangent to the graph of $f_p$. You can try implicit differentiation. Start with $$ y = \frac{9\sqrt{x^2 + p}}{x^2 + 2}. $$ Multiply by $x^2 + 2$ to get $$ y(x^2 + 2) = 9\sqrt{x^2 + p}. $$ Squaring, you get $$ y^2 (x^2 + 2)^2 = 81 (x^2 + p). $$ Differentiate both sides implicitly by $x$: $$ 2yy'(x^2 + 2)^2 + y^2 2 (x^2 + 2) 2x = 162x. $$ Now, plug in all the data ($x = -1$, $y'(-1) = 2.5$) to get a quadratic equation for $y(-1) = y$: $$ -12y^2 + 45y = -162. $$ Solutions are $y = 6$ and $y = \frac{-9}{4}$, but notice your function is always positive, so $y(-1) = 6$ and the line is $2.5x + 8.5$.


The point $A$ is $(x, \space f_p(x)) = \left(x, \space \dfrac{9\sqrt{x^2+p}}{x^2+2}\right)$ at $x = -1$ this gives $A = \left(-1, \space 3\sqrt{1+p}\right)$.

The line $k$ with a slope of $2.5$ touches $f_p$ in $A$ i.e passes through point $A$ then $k =mx+c$ where $m=2.5$ Then you substitute point $A$ in the equation to get the constant $c$, that is $$3\sqrt{1+p} = 2.5 \times -1 + c \quad \Rightarrow \quad c= 3\sqrt{1+p}+2.5$$ which implies that $$k = 2.5x +3\sqrt{1+p}+2.5$$

If by touch you meant tangent to $f_p$ then the only way to find the value of $p$ is to find the derivative of $f_p$.

Instead of using the quotient rule, one can also apply the product rule on $$f_p(x) = 9(x^2+p)^{1/2}(x^2+2)^{-1}$$ which gives

$\quad \quad \quad \cfrac {df_p}{dx} = 9 \left[ ((x^2 +p)^{1/2})' \cdot (x^2+2)^{-1} + (x^2 +p)^{1/2} \cdot ((x^2+2)^{-1})'\right ]$

$\quad \quad \quad \quad \quad = 9 \left[ x(x^2 +p)^{-1/2} \cdot (x^2+2)^{-1}\space -\space 2x (x^2 +p)^{1/2} \cdot (x^2+2)^{-2}\right ]$

Compute $f'_p(-1)$ easily as

$\quad \quad \quad =9 \left[ \left ((-1)((-1)^2 +p)^{-1/2} \cdot ((-1)^2+2)^{-1}\right ) + \left (-2(-1) ((-1)^2 +p)^{1/2} \cdot ((-1)^2+2)^{-2}\right)\right ]$

$\quad \quad \quad =9 \left[ -\cfrac 13(1 +p)^{-1/2}+\cfrac 29 (1 +p)^{1/2} \right ]$ $\quad =\quad-3(1 +p)^{-1/2}+2(1 +p)^{1/2} = 2.5$

Let $X = 1+p$, then $\cfrac {-3}{\sqrt X}+2\sqrt X = 2.5$. Solving this equation gives $X = 1+p = 4 \quad \Rightarrow \quad p =3$ and $$k = 2.5x +8.5$$

  • $\begingroup$ This certainly is elegant! $\endgroup$ – JohnPhteven Oct 28 '12 at 22:39

I see that you're looking for something concise, but I think that you'll find this to be the most clear and failsafe.

I find that trying to simplify things as much as possible generally helps.

To start off, note what the problem is trying to ask. We have a fairly steep line, and it meets the function to the left of the $x$-axis. We want this line.

The Derivative

Now, we need the derivative. There's usually a fairly straightforward method to solving many of the questions that you'll come across, so let's see if we can simplify things... We have the equation:


I note the "$x^2 + \text{something}$" term in both the numerator and denominator. We can use that as a good starting point, and break up the problem in pieces using this term as the basic building block:


This seems more managable to me. The key for derivatives, and later on calculus if you learn that, is trying to break things into more manageable pieces. We can pretty much ignore the 9 in the equation, so let's try to write this equation even more simply:


This fits into the product rule for derivatives:

$$\frac{d}{dx}u\cdot v=u\frac{dv}{dx}+v\frac{du}{dx}$$

Once again, we have:

$$\frac{1}{9}f_p(x)=\underbrace{\left(\sqrt{x^2+p}\right)}_\bf{u} \underbrace{\cdot\frac{1}{x^2+2}}_\bf{v}$$

So now to do this entire derivative, we really only have to calculate $du$ and $dv$. So:

$$u = \sqrt{x^2+p} = (x^2+p)^{1/2}\implies du = 1/2(x^2+p)^{-1/2} \cdot 2x$$ $$v = \frac{1}{x^2+2} = (x^2 + 2)^{-1} \implies dv = -1(x^2 + 2)^{-2} \cdot 2x$$

In the derivatives above, I used a dot to seperate out the pieces of the derivative again. It's really helpful when there are complicated equations involved. That little dot helps simplify things by breaking them up into pieces, so that the equations or wahtever become much more manageable. It can really help to cut down on mistakes by breaking things up into more manageable pieces.

Now that those pieces are done (and hopefully doublechecked) we can piece together the original derivative. At this time, we could take an additional step, and try to simplify the pieces:

$$du = 1/2(x^2+p)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+p}}$$

$$dv = -1(x^2 + 2)^{-2} \cdot 2x = \frac{-x}{(x^2 + 2)^2}$$

Remembering the product rule, we have:

$$u\frac{dv}{dx} + v\frac{du}{dx} = \left(\sqrt{x^2+p}\frac{-x}{(x^2 + 2)^2}\right) + \left(\frac{1}{x^2+2} \frac{x}{\sqrt{x^2+p}}\right)$$

Do you remember the 9 that we factored out? We can now bring this back. Besides that, nothing else needs to be done with the derivative except to maybe simplify it.

The Equation for the Line

Closing Remarks I have found that even for simple questions/problems, and especially for complicated and tediuous questions/problems, that mathematics software really helps. I know that it may be hard to get, and that you may not be able to always use it, but it really helps to explore math. I've found that "playing" with math is a great way to learn and understand things. That and practice (with random/different problems) can make a big difference in math ability.

  • $\begingroup$ isn't this just a remix of my solution? $\endgroup$ – user31280 Oct 28 '12 at 11:16
  • $\begingroup$ @F'OlaYinka: In a way, yes. But I thought that my presentation may be easier to digest. I tried to stress simplification as a general approach. I also tried to stress the system of doing this, instead of the equations for this particular problem. I'm sorry that this may seem a rehash of your equations, but I thought that stressing the overall system of doing things was more critical than the particulars of the problem. You start out with a bunch of equations that already hint at the solution, while I take more time to introduce the equation of the line. Please forgive me for similarity. $\endgroup$ – Matt Groff Oct 28 '12 at 12:23
  • $\begingroup$ @F'OlaYinka: Let's hope that the OP gets a better understanding of the approach. Maybe if the OP sees that our approaches are similar, he/she may decide that we have a good approach that works for alot of mathematicians. If you still think that my approach is too similar, I can remove it. I just would really like to see the system work for the OP! I'm open to suggestions, though. Perhaps we could combine our solutions... I didn't see the product rule presented in your solution, which seems to work really well for me. I would really like to see that in your solution. $\endgroup$ – Matt Groff Oct 28 '12 at 12:33
  • $\begingroup$ This indeed is easier than my method $\endgroup$ – JohnPhteven Oct 28 '12 at 19:05
  • $\begingroup$ @MattGroff I am the OP $\endgroup$ – JohnPhteven Oct 28 '12 at 22:34

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