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What does $\frac{dg}{dx}$ mean?

Specifically, I'm trying to solve$$ \frac{1}{3}\frac{dg}{dx}\frac{1}{1+g^2} $$ where $$ g(x) = \frac{3x\left(1-x^2\right)}{x^4-4x^2+1} $$

I know $\frac{d}{dx}$ just means differentiate with respect to $x$ but I have no idea what this $\frac{dg}{dx}$ business is.

Cookie
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chriskinda
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    "Solve" is not the right word here. It often gets used as a catch-all term by people who don't know what word to use, but it is not correct to do that. "Evaluate" fits better. – Michael Hardy Mar 02 '15 at 04:30
  • @MichaelHardy Do you say that because I'm not trying to find a specific number? – chriskinda Mar 02 '15 at 04:36

2 Answers2

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Leibniz's notation $\frac{dg}{dx}$ represents the derivative of your function $g(x)$. You can write this as $$\frac{dg}{dx}=\frac{d}{dx} g(x).$$ As you know that $\frac{d}{dx}$ means, then you might know that $\frac{d}{dx} g(x)$ means "differentiate the function $g(x)$ with respect to $x$".

Cookie
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  • The standard word is "differentiate". Admittedly that is something of a linguistic anomaly, but nonetheless it is standard. – Michael Hardy Mar 02 '15 at 04:31
  • @MichaelHardy I know. I was trying to quote what the OP said with a little bit of variation, hopefully to connect better with the OP. :) I do understand that the word "derive" is rather vague and might be confused with its lexical definition in a non-mathematical context. – Cookie Mar 02 '15 at 04:32
  • So you're saying I should take $\frac{d}{dx} \frac{1}{1+(\frac{(3x(1-x^2)}{x^4-4x^2+1})^2}$? Because when I put this in Mathematica I do not get the answer I'm looking for. – chriskinda Mar 02 '15 at 05:23
  • No, you should take $\frac d{dx} \frac{3x\left(1-x^2\right)}{x^4-4x^2+1}$. I do not use Mathematica, but WolframAlpha says that this derivative (http://www.wolframalpha.com/input/?i=derivative+of+g%28x%29+%3D+%5Cfrac%7B3x%5Cleft%281-x%5E2%5Cright%29%7D%7Bx%5E4-4x%5E2%2B1%7D) is $$\frac{dg}{dx}=\frac{3(x^6+x^4+x^2+1)}{(x^4-4x^2+1)^2}.$$ – Cookie Mar 02 '15 at 05:26
  • @dragon You'd need to multiple by $\frac{1}{3}$ per the initial equation but yes I get that too. Can you go to this link and verify that the answer that was presented is correct? http://math.stackexchange.com/questions/1170993/fracddx-frac13arctan-frac3x1-x2x4-4x21c-fracx41x – chriskinda Mar 02 '15 at 05:42
  • @Dragon Please? It is the first part of this question and the reason I don't like the answer you've presented but I agree with. :) – chriskinda Mar 02 '15 at 20:27
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Do you mean $\frac{1}{3}\frac{dg}{dx}\frac{1}{1+g^2}$ or $\frac{1}{3}\frac{d}{dx}\frac{g}{1+g^2}$?

In the first case, you compute $\frac{d}{dx}\frac{3x(1 - x^2)}{x^4 - 4x^2 + 1}$ and multiply the answer by $\frac{1}{1+g^2}$.

In the second case, you use the chain rule to compute $\frac{1}{3}\left(\frac{d}{dg}\frac{g}{1+g^2}\right) \frac{dg}{dx}$

I just want to make sure that you are aware there is a difference.

Steven Alexis Gregory
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