Blog Post 73
by EpicSkills32, Mar 16, 2014, 5:36 AM
![$\ [\text{Blog Post 73}] $](http://latex.artofproblemsolving.com/0/c/2/0c27efb05bf3f08d44a56fe055f05c429dafa92d.png)
**NVM ABOUT EDITING**
**IT WAS ACTUALLY CORRECT**
Regarding my previous post, I'm surprised that csmath was the only one who commented. It turns out that actually both were extremely easy to calculate.
csmath is right about the second one.
![\[ \dfrac{d}{dx} \left[\dfrac{1}{3x^2}\right] \]](http://latex.artofproblemsolving.com/1/a/e/1aed22cdb4c8131d95834dfd8948c5376a710400.png)
...is extremely easy to calculate. As csmath suggested, you can just write the fraction as a term with a negative exponent, and then chain rule and...
But wait. There is an extremely useful identity that we can use:
![\[ \dfrac{d}{dx}\left[\dfrac{1}{f(x)}\right]=\dfrac{-\dfrac{d}{dx}[f(x)]}{(f(x))^2} \]](http://latex.artofproblemsolving.com/2/8/5/285603714aaea0b398b859db48430bf2beddd5b9.png)
We apply this now:
![\[ \dfrac{d}{dx}\left[\dfrac{1}{3x^2}\right]=\dfrac{-\dfrac{d}{dx}\left[3x^2\right]}{\left(3x^2\right)^2} \]](http://latex.artofproblemsolving.com/1/d/5/1d59ee9732113778f3b1fe19e74651714ec609be.png)
![\[ = \dfrac{-6x}{9x^4} \]](http://latex.artofproblemsolving.com/7/4/f/74fb47c0286a465a8a622eaa0695fdfd66568b1d.png)
![\[ =\boxed{\dfrac{-2}{3x^3}} \]](http://latex.artofproblemsolving.com/0/d/3/0d3835312aa9da1f17854f3a651f979f0702b211.png)
And we're done.
The first one that I had looks a bit harder, but it's actually not.
The problem was:
Calculate the derivative of:
![\[ \dfrac{1}{1-\dfrac{1}{x^2}} \cdot \dfrac{2}{x^3} \]](http://latex.artofproblemsolving.com/6/a/a/6aa5b28d23d0b975902541d0f4c92d9352f8e285.png)
At first glance, it looks pretty nasty (especially the first term), but let’s see if algebraic simplification will…uh….simplify anything.
With the first term, we first simplify the fraction:
![\[ \dfrac{1}{1-\dfrac{1}{x^2}} = \dfrac{x^2}{x^2-1} \]](http://latex.artofproblemsolving.com/e/b/b/ebb4964f380519cc0ff760ef22faf78659217142.png)
We use the quotient rule to find the derivative:
![\[ \dfrac{d}{dx}\left[\dfrac{x^2}{x^2-1}\right]=\dfrac{\left(x^2-1\right)(2x)-\left(x^2\right)(2x)}{\left(x^2-1\right)^2} \]](http://latex.artofproblemsolving.com/7/2/e/72e8152c4c0ff4cee24edb4fd6188953942147fb.png)
![\[ = -\dfrac{2x}{\left(x^2-1\right)^2} \]](http://latex.artofproblemsolving.com/7/b/8/7b82abce0ab7a1209ac1d1d53069cb7500f6059d.png)
Now we need to worry about the
Unfortunately, we can't apply the reciprocal rule here, since there's a 2 in the numerator. Hm......or maybe we can.....
![\[ \dfrac{2}{x^3}=\dfrac{1}{\dfrac{1}{2}x^3} \]](http://latex.artofproblemsolving.com/c/8/b/c8b8d044db48c109c7429cbec82ff2658fee1e4e.png)
Well whaddaya know....it's in a form we can apply the rule on!
Using our handy-dandy rule, we get:
![\[ \dfrac{d}{dx}\left[\dfrac{2}{x^3}\right]=\dfrac{d}{dx}\left[\dfrac{1}{\dfrac{1}{2}x^3}\right] \]](http://latex.artofproblemsolving.com/8/2/8/828e9373aba6805575525a4f23abcfea61981fc0.png)
![\[ = \dfrac{-\dfrac{3}{2}x^2}{\dfrac{1}{4}x^6} \]](http://latex.artofproblemsolving.com/d/5/d/d5d10693829660b8f60064135677debfe466413a.png)
![\[ = -\dfrac{12x^2}{2x^6} = -\dfrac{6}{x^4} \]](http://latex.artofproblemsolving.com/c/c/f/ccfcf6144d18c1ded49a1fc2ebe80aa46e601ede.png)
Since we have the derivative of a product, we have to remember to multiply by the functions themselves alternately, since the derivative of a product is not just the product of the derivatives.
![\[ \dfrac{d}{dx}\left[\dfrac{1}{1-\dfrac{1}{x^2}} \cdot \dfrac{2}{x^3}\right] \]](http://latex.artofproblemsolving.com/4/7/4/4747637aca9e8a3c610f2a309efc9c62a121162b.png)
would be the derivative of the first term times the second term, plus the derivative of the second times the first term.
Which would be:
Simplifying:
\[ = \dfrac{-4x}{x^3\left(x^2-1\right)^2} + \dfrac{-6}{x^4\left(1-\dfrac{1}{x^2}\right)} = \dfrac{-4-6\left(x^2-1\right)}{x^2\left(x^2-1)^2} \]![\[ = \dfrac{-2\left(3x^2-1\right)}{x^2\left(x^2-1\right)^2} \]](http://latex.artofproblemsolving.com/2/6/8/268db141c06913dc05aa01482fef687b6146c5bc.png)
And there we have it. Our final answer would be:
![\[ \dfrac{d}{dx}\left[ \dfrac{1}{1-\dfrac{1}{x^2}} \cdot \dfrac{2}{x^3} \right] =\dfrac{-2\left(3x^2-1\right)}{x^2\left(x^2-1\right)^2} \]](http://latex.artofproblemsolving.com/8/2/8/8282cd8568d4329e254dfcdbdc6a295c2b91fb76.png)
This post has been edited 5 times. Last edited by EpicSkills32, Mar 18, 2014, 4:39 AM
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