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Ok, I am going through the MIT Open Course ware course on single variable calculus. I have never taken calculus before, so I apologize of this is a really trivial question.

I know that with $f(x)=x^n$ then f'(x)=nx^{n-1}.

Without using this trick and fully working it out I can't seem to come to the derivative of $f(x)=80-5x^2$.

I basically boil it down to $(-5/dx)(x+dx)(x+dx)$. I cannot seem to get $dx$ out of the denominator so that I don't have a division by zero as $dx$ tends to 0... Am I brain farting something here or is this why I should just skip to the aforementioned shortcut?

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    Disregard how I got to $(−5/dx)(x+dx)(x+dx)$ as I was making a very foolish mistake which I have noted below...2011-12-28

4 Answers 4

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Here is the simplest method that I would try to use:

$ f(x)=80-5x^{2} \, \implies \, f'(x)=(0)+\Big((-5) \times (2) \times x^{(2)-(1)} \Big) \ \\ f'(x)=0+(-10 \times x^{1} ) \ \\ f'(x)=-10x \ $

  • I've made an edit to correct my answering post. I think it is OK now.
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    I see now. That IS very good... :) Thanks for both editing and noticing. :)2012-05-03
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One way of doing this is directly from the definition of derivative as a limit: f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}. In your case, f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{(80-5(x+h)^2)-(80-5x^2)}{h}. Now you just need to first expand and then simplify the numerator to calculate the limit.

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    Right, I was screwing up the expansion and not keeping a negative sign around... Sigh. How do you use the fancy fonts and symbols?2011-12-28
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If you're going to do it using fractions involving $dx$, then you need $ -5\frac{(x+dx)^2-x^2}{dx}. $ After expanding and doing routine cancelations, you then discard the infinitesimal part $dx$.

The more modern way is to look at $ -5\frac{(x+\Delta x)^2-x^2}{\Delta x} $ and find the limit as $\Delta x$ approaches $0$.

So first do a bit of algebra: $ -5\frac{(x+\Delta x)^2-x^2}{\Delta x} = -5\frac{x^2 + 2x\;\Delta x + (\Delta x)^2 - x^2}{\Delta x} = -5\frac{(2x+\Delta x)\Delta x}{\Delta x} = -5(2x+\Delta x). $ As $\Delta x$ approaches $0$, this approaches $-10x$ .

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    Furthermore, I just wanted to add you taking the time to fully work through this problem is precisely what showed me my mistake. If I could, I would up-vote this answer and I plan on doing so whenever it is I get the amount of points to unlock that ability.2011-12-28
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This trick works but you must remember that the derivative of the sum of two functions is the sum of their derivatives, that the derivative of a constant is $0$, and also that the derivative of the product of a constant and a function is equal to the product of the constant and the derivative of the function. So if:

$f(x) = g(x) + h(x)$

$g(x) = 80$

$h(x) = -5x^2$

then:

h'(x) = -10x

g'(x) = 0

f'(x) = g'(x) + h'(x) = -10x.