# Rate of Change/ derivative question

• Jul 28th 2013, 06:37 AM
Welkin
Rate of Change/ derivative question
Still playing catch up with my basic algebra so I've written out another question similar to one I'm doing for an assignment and it'd be great if anyone could have a look over it and see just how many errors I'm making.

$\displaystyle f(x) = 3x(1-x)s^4$

$\displaystyle \frac{df}{dx} = \frac{\lim}{\Delta x \rightarrow \infinity} = \frac{f(x+ \Delta x) - f(x)}{\Delta x}$

$\displaystyle 3(x + \Delta x) (1-(x + \Delta x))s^4 - 3x(1 - x) s^4$

$\displaystyle 3x + 3 \Delta x +1 -x - \Delta x - 3x + 3x^2$

$\displaystyle 3 \Delta x +1 -x - \Delta x - 3 \Delta x + 3x^2$

* Can I cancel out the $\displaystyle s^4$ and $\displaystyle -s^4$ here or do they need to be multiplied (factored out?) with their respective brackets?

$\displaystyle 3 \Delta x (-x - \Delta x) + 3x^2$

$\displaystyle \frac{3 \Delta x +1 -x - \Delta x +3x^2}{ \Delta x}$

$\displaystyle \frac{\lim}{\Delta x \rightarrow \infinity} = 1 - x + 3x^2$
• Jul 28th 2013, 06:45 AM
HallsofIvy
Re: Rate of Change/ derivative question
Are you saying you do not know what "canceling" means? If we have a fraction like $\displaystyle \frac{s}{s}$ then, yes, we can "cancel" because we know that fraction is "1". If we have, say, $\displaystyle \frac{as}{bs}$, we know we can write it as $\displaystyle \left(\frac{a}{b}\right)\left(\frac{s}{s}\right)$ which is then equal to $\displaystyle \frac{a}{b}$. If we have something like $\displaystyle \frac{a+ s}{b+ s}$ we cannot cancel because that is NOT equal to $\displaystyle \frac{a}{b}+ \frac{s}{s}$.

Similarly, s+ (-s)= 0 so we can "cancel" when we have such added. We can cancel, say, (a+ s)+ (b- s) because that can be written as (a+ b)+ (s- s). However, as- bs cannot be written that way and we cannot cancel in that way.
• Jul 28th 2013, 06:58 AM
Welkin
Re: Rate of Change/ derivative question
So if the first step in setting out the problem is:

$\displaystyle \frac {3(x + \Delta x) (1-(x + \Delta x))s^4 - 3x(1 - x) s^4}{\Delta x}$

To keep it simple if I just take the second part $\displaystyle f (x)$ which is $\displaystyle - 3x(1 - x) s^4$

Should that be expanded as:

$\displaystyle (-3x + 3x^2)s^4$

$\displaystyle -3xs^4 - 3x^2s^4$
• Jul 28th 2013, 08:42 PM
Welkin
Re: Rate of Change/ derivative question
Ok I've had another try from scratch, would be good to know if it's any closer to the right process or if I'm just switching one lot of errors for another.

$\displaystyle f(x) = 3x(1-x)s^4$

Where h = $\displaystyle \Delta x$ (just finding it easier to keep it clear when using pen and paper)

$\displaystyle \frac{df}{dx} = \frac{\lim}{h \rightarrow 0} = \frac{f(x+ h) - f(x)}{h}$

$\displaystyle \frac {3(x+h) (1- (x+h)s^4 - 3x(1-x)s^4}{h}$

$\displaystyle \frac{(3x +3h)(1-x-h)s^4 - 3xs^4 +3x^2s^4}{h}$

$\displaystyle \frac{(3x-3x^2-3xh+3h-3hx-3h^2)s^4 - 3xs^4 +3x^2s^4}{h}$

$\displaystyle \frac{3xs^4-3x^2s^4-3xhs^4+3hs^4-3hxs^4-3h^2s^4 - 3xs^4 +3x^2s^4}{h}$

$\displaystyle \frac{-3xhs^4+3hs^4-3hxs^4-3h^2s^4}{h}$

$\displaystyle \frac{-6xhs^4+3hs^4-3h^2s^4}{h}$

$\displaystyle -6xs^4+3s^4-3hs^4$

$\displaystyle = \frac{\lim}{h \rightarrow 0} -6xs^4 + 3s^4$
• Jul 29th 2013, 11:09 PM
Welkin
Re: Rate of Change/ derivative question
Well this serves me right for not wanting to post an actual test question, the original function was similar but $\displaystyle s^4$ was actually $\displaystyle m^3$ as in meters cubed and yeah the f(x) should have been a hint there. No wonder the problem looked a little longer then what I had been practicing (I'm doing more of a pre uni course and it was the two variables that made me want to check if I was on the right track). Anyway it'd still be good if anyone can let me know if the expansion and simplification of the top line of the equation seems right (in the second attempt) as I think it's the algebra more than the concepts that are causing me problems at the moment..... well that and reading.