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I am suppose to differentiate $$x^3 -4x +6,$$ but I can not get the correct answer. I get $\frac{\mathrm{d} }{\mathrm{d} x}(x^3) = 3x^2$ using the power rule and then $\frac{\mathrm{d} }{\mathrm{d} x}(4x) = 4$ and $\frac{\mathrm{d} }{\mathrm{d} x}(6) = 1$. This is not correct but I do not know why.

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    $\begingroup$ Sloppy notation! $x^3$ is not the same thing as $3x^2$, so you mustn't write $x^3=3x^2$. It's the derivative of $x^3$ that's $3x^2$, so you have to write $D(x^3)=3x^2$ or $\frac{d}{dx}(x^3)=3x^2$, or something similar (depending on what notation for derivatives that you prefer). $\endgroup$ – Hans Lundmark Sep 17 '11 at 19:59
  • $\begingroup$ Sorry I didn't know how to do that with latex so I just thought it was assumed. $\endgroup$ – user138246 Sep 17 '11 at 20:03
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    $\begingroup$ @Jordan: Well, obviously we understood what you meant, so we'll forgive you this time. But please don't make it into a habit! ;-) $\endgroup$ – Hans Lundmark Sep 17 '11 at 20:05
  • $\begingroup$ @Jordan Carlyon:I did it this time,you can now check the edit :) $\endgroup$ – Quixotic Sep 17 '11 at 20:08
  • $\begingroup$ Some things I tried to emphasize when I was teaching Calculus: (1) In high-school Algebra, you don't make a distinction between letters and explicit numbers: $a$ behaves the same as $2$. But in Calculus, we worry very much about whether a letter represents a constant or a variable, because they’re treated entirely differently in Calculus. (2) Don’t ever put an equals sign between two things that are not equal. $\endgroup$ – Lubin Apr 23 '12 at 17:48
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Everything is correct, except that the derivative of a constant (like 6) is always 0.

You can still see this fact from the power rule. Write 6 as $6x^0$. The power rule says that the derivative is $6 \cdot 0 x^{-1}$, which is 0.

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  • $\begingroup$ But isn't $6^0$ = 1? Also how does a more complex number like $4x^3$ fit into the rule? How does x being 4x affect the problem? $\endgroup$ – user138246 Sep 17 '11 at 19:56
  • $\begingroup$ @Jordan The exponent is only on the $x$ in my example, which is why it works. $6(x^0) = 6(1) = 6$ $\endgroup$ – Austin Mohr Sep 17 '11 at 19:58
  • $\begingroup$ @Jordan A constant multiple can be "ignored" until the end. To find the derivative of $4x^3$, find the derivative of $x^3$ first and then multiply by 4. $\endgroup$ – Austin Mohr Sep 17 '11 at 20:04
  • $\begingroup$ Ok that makes sense, thank you. $\endgroup$ – user138246 Sep 17 '11 at 20:07
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    $\begingroup$ Also don't forget to get a feeling for thiss. The derivative of a constant is zero because no matter how we change x (which is represented by $dx$ in a not so trivial way) the resulting change in the constant is ... zero because it is a constant so it doesn't change. Hence you don't need to memorize it. You need to see it and get used to it. $\endgroup$ – user13838 Sep 17 '11 at 23:19
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You nearly got everything correct, only the derivative of the constant is wrong. However, you should watch out with your notation, you don't write $x^3 = 3x^2$, instead you write either $\frac{d}{dx}x^3 = 3x^2$ or $(x^3)^{'} = 3x^2$

Also remember that when you differentiate $x^3-4x+6$ you first have to say that you can derive $x^3$, $4x$ and $6$ separately, since the derivate of a sum equals the sum of the derivate.

Thus $\frac{d}{dx}(x^3-4x+6) = 3x^2 -4$.

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  • $\begingroup$ Yep, typo. Thanks for pointing it out. $\endgroup$ – sxd Sep 17 '11 at 20:06
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Use the rule of the sum of functions, $$\frac d{dx}(x^3-4x+6)=\frac d{dx}(x^4)-\frac d{dx}(4x)+\frac d{dx}(6)=\underline{3x^2-4}\\\text{NOT }\;\; 3x^2-4 \;\;\underline{+ 6} $$

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  • $\begingroup$ Thanks for the timely edit, Bidit. $\endgroup$ – The Chaz 2.0 Apr 23 '12 at 17:19
  • $\begingroup$ @TheChaz Anytime :) $\endgroup$ – funktor Apr 23 '12 at 17:32

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