# Derivative of $f(x)=80-5x^2$

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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Can you write out the work you did to get that result? – tomcuchta Dec 28 '11 at 3:09
How much "trick" do you want to avoid? Is the product rule fair game, for example? – Henning Makholm Dec 28 '11 at 3:13
How did you come up with $(-5/dx)(x+dx)(x+dx)$? – Jack Dec 28 '11 at 3:55
Disregard how I got to $(−5/dx)(x+dx)(x+dx)$ as I was making a very foolish mistake which I have noted below... – isolier Dec 28 '11 at 5:48

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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Wow, every time I did this I forgot to keep the negative sign on the $-x^2$ so that when I expanded the $(x+dx)^2$ the $x^2$ and the $-x^2$ didn't cancel and led me completely astray. I am both relieved you have helped me through this and immensely annoyed at myself... Thanks man! – isolier Dec 28 '11 at 5:41
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. – isolier Dec 28 '11 at 5:51

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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The symbol $\iff$ means "if and only if". It is a statement of logical equivalence. While $f(x)=80-5x^2$ does imply that $f'(x) = 0 + (-5)(2)x^{2-1}$, the converse does not hold, so $\implies$ is justified, but $\impliedby$ is not. – Arturo Magidin May 3 '12 at 19:46
So, I should use $\, \implies \,$ instead of $\, \iff \,$ for this answering post...? But, what changes @rschwieb have been made for my answer post, I could not recognize anything different or new since my original post...? Thanks for the noticing by the way. – Kerim Atasoy May 3 '12 at 20:31
You can see the changes by clicking on the link after edited. rschwieb corrected your grammar ("most simpliest" --> "simplest"; the superlative 'est' already has an implied 'most'; and "to do it"-->"to use") – Arturo Magidin May 3 '12 at 20:40
I see now. That IS very good... :) Thanks for both editing and noticing. :) – Kerim Atasoy May 3 '12 at 20:47

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:

$h'(x) = -10x$

$g'(x) = 0$

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

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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? – isolier Dec 28 '11 at 5:42