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I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$ I simplify this a little bit, by moving the constant multiplicator out of the derivative: $$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(\frac{x^2-6x-9}{x^2(x+3)^2}\right) $$ But, using the quotient-rule, the resulting expressions really get unwieldy: $$ \frac{1}{2} \frac{(2x-6)(x^2(x+3)^2) -(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} $$

I came up with two approaches (3 maybe):

  1. Split the terms up like this: $$ \frac{1}{2}\left( \frac{(2x-6)(x^2(x+3)^2)}{(x^2(x+3)^2)^2} - \frac{(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} \right) $$ so that I can simplify the left term to $$ \frac{2x-6}{x^2(x+3)^2}. $$ Taking this approach the right term still doesn't simplify nicely, and I struggle to combine the two terms into one fraction at the end.

  2. The brute-force-method. Just expand all the expressions in numerator and denominator, and add/subtract monomials of the same order. This definitely works, but i feel like a stupid robot doing this.

  3. The unofficial third-method. Grab a calculator, or computer-algebra-program and let it do the hard work.

Is there any strategy apart from my mentioned ones? Am I missing something in my first approach which would make the process go more smoothly? I am looking for general tips to tackle polynomial fractions such as this one, not a plain answer to this specific problem.

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    $\begingroup$ Frankly, I think that we waste a lot of time worrying about simplifying things like this. Try a couple of things. If you can't get something nice relatively quickly, throw a CAS at it and get on with your life. In the real world, you will encounter very few situations where you want to factor a polynomial and are also able to do so (e.g. almost every polynomial of degree $5$ or higher cannot be factored; in many real-world settings, the polynomials come pre-factored; etc). $\endgroup$ – Xander Henderson Jun 6 '20 at 2:51
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    $\begingroup$ In examples like this (that is, if you want to differentiate a rational function), you might save some time using logarithmic differentiation. $\endgroup$ – Xander Henderson Jun 6 '20 at 2:52
  • $\begingroup$ @XanderHenderson I agree that in modern times one should focus more on understanding and applying concepts, than on pure computation. However, in this case i am happy to have been pointed to logarithmic differentiation, partial fractions and polynomial long division which made me realize some knowledge-gaps i was unaware of, which would not have happened had i used a CAS. $\endgroup$ – LeonTheProfessional Jun 6 '20 at 7:16
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Logarithmic differentiation can also be used to avoid long quotient rules. Take the natural logarithm of both sides of the equation then differentiate: $$\frac{y'}{y}=2\left(\frac{1}{x-3}-\frac{1}{x}-\frac{1}{x+3}\right)$$ $$\frac{y'}{y}=-\frac{2\left(x^2-6x-9\right)}{x(x+3)(x-3)}$$ Then multiply both sides by $y$: $$y'=-\frac{{\left(x-3\right)}^3}{x^3{\left(x+3\right)}^3}$$

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    $\begingroup$ Great!! Yet another approach! I will look into logarithmic differentiation a bit more. With these tools i don't need to feel like a stupid robot anymore ;) $\endgroup$ – LeonTheProfessional Jun 5 '20 at 17:41
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    $\begingroup$ Yep. You should logarithmic differentiation because it's more applicable than the other methods previously mentioned. Especially if there are square root, cube root, etc. of polynomials in the numerator/denominator. $\endgroup$ – Ty. Jun 5 '20 at 17:43
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HINT

To begin with, notice that \begin{align*} x^{2} - 6x - 9 = 2x^{2} - (x^{2} + 6x + 9) = 2x^{2} - (x+3)^{2} \end{align*} Thus it results that \begin{align*} \frac{x^{2} - 6x - 9}{2x^{2}(x+3)^{2}} = \frac{2x^{2} - (x+3)^{2}}{2x^{2}(x+3)^{2}} = \frac{1}{(x+3)^{2}} - \frac{1}{2x^{2}} \end{align*}

In the general case, polynomial long division and the partial fraction method would suffice to solve this kind of problem.

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  • $\begingroup$ Thank you! I think i stared too long at this problem, and have to take a step back to notice these patterns. I will examine the problem again (tomorrow) with your hints, and see if any more questions arise. If so, i will comment here. If no other answers of overwhelming enlightment appear, i will mark this answer as the accepted one. :) $\endgroup$ – LeonTheProfessional Jun 5 '20 at 17:35
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Note that $x^2-6x-9 = (x-3)^2 - 18$. So after pulling out the factor of $\frac 12$, it suffices to compute $$\frac{d}{dx} \left(\frac{x-3}{x(x+3)}\right)^2$$ and $$\frac{d}{dx} \left(\frac{1}{x(x+3)}\right)^2.$$ These obviously only require finding the derivative of what's inside, since the derivative of $(f(x))^2$ is $2f(x)f'(x)$.

For a final simplification, note that $$\frac{1}{x(x+3)} = \frac{1}{3} \left(\frac 1x - \frac{1}{x+3}\right),$$ so you'll only ever need to take derivatives of $\frac 1x$ and $\frac {1}{x+3}$ to finish, since the $x-3$ in the numerator of the first fraction will simplify with these to give an integer plus multiples of these terms.

As a general rule, partial fractions will greatly simplify the work required in similar problems.

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  • $\begingroup$ I found this partial fraction using polynomial long division, yet it didn't appear to me, that any remainder appearing there would obviously disappear when taking the derivative. This really is a great hint! $\endgroup$ – LeonTheProfessional Jun 5 '20 at 17:39
  • $\begingroup$ @hdighfan I think there is a slight typo - it should read that the derivative of $\left(f(x)\right)^2$ is $2f(x)f'(x)$. $\endgroup$ – Zubin Mukerjee Jun 6 '20 at 3:01
  • $\begingroup$ I have edited it to fix, please revert if not wanted $\endgroup$ – Zubin Mukerjee Jun 6 '20 at 20:18
  • $\begingroup$ Ah, of course. Thanks for the edit. $\endgroup$ – hdighfan Jun 6 '20 at 20:19

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