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?
 A: 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. 
A: 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 $-10x$
.
A: 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:
$h'(x) = -10x$
$g'(x) = 0$
$f'(x) = g'(x) + h'(x) = -10x$.
A: 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.

