Differentiation

Ah so was the answer: 0x^-1
Thanks very much...

Well, the answer is 0. Because 0 * anything = 0.

Ohh thanks. Will they ask these type of questions on IGCSE too?

Depends what IGCSE.

Anyways, can you now do the original question in your first post? Please do it step-by-step like I did in my previous post! What do you get?

So dy/dx = 3*2x^3-1

dy/dx = 6x^2 ?

We want to differentiate $y = 2x^3 + x^2 - 8x + 3$.

So, you've correctly differentiated the first term, well done. Now what about the rest?

I don't know the rest this is hard

Why aren't you doing it step-by-step like I told you do?

If we want to differentiate the sum of all those things, we can just differentiate each term separately and then add them up.

SO, to differentiate $y = 2x^3 + x^2 - 8x + 3$, you need to be able to differentiate each bit seperately:

(i) differentiate 2x^3, you've done this correctly.
(ii) differentiate x^2, you can do this, c'mon!
(iii) differentiate -8x, you can do this!
(iv) differentiate 3, what did we say the derivative of a constant is?

Oh I think I can do it and then do I put a + sign in between them all

Yep - pretty much. What do you get?

By the way, you might want to watch this as a very detailed example. Please take some time to sit down and understand all of this.

Thanks very much - I'm very tired atm so cannot concentrate very well atm.

Come back and read this thread again in the morning.

Ah so was the answer: 0x^-1
Thanks very much...

Okay, here are some rules that you should get used to for differentiating. What $\frac{dy}{dx}$ is is a linear operator. It's read as "the derivative of y with respect to x". Because it's linear, it's a very easy and nice operator.

So, we have, intuitively:



All this is saying that if you want to differentiate the sum of two (or more!) terms, you can differentiate each term separately and then add them together.

So, for example - you know that when you differentiate $x^1$, you get $1x^0 = 1$. So if you were to differentiate $2x$ you can simply write $2x = x+x$ and then it follows that $\frac{d}{dx}(2x) = \frac{\mathrm{d}}{\mathrm{d}x}(x) + \frac{\mathrm{d}}{\mathrm{d}x}(x) = 1 + 1 = 2$. Which makes sense!

Following on from that rule, it follows just as intuitively that

.

That is, if you want to differentiate a constant multiple of $x$, you can pull the multiple out, differentiate the x term and then put the constant back in. This makes total sense for differentiating $2x$ because we can do $\frac{\mathrm{d}}{\mathrm{d}x}(2x) = 2\frac{\mathrm{d}}{\mathrm{d}x}(x) = 2 (1) = 2$ just like last time!

Now, you've just learnt that if you want to differentiate $x$ raised to a power, you do this:



i.e: all you need to do is multiply the entire thing by the power and then subtract 1 from the power. Do you understand this? Please tell me if you don't!

With these three rules, you can differentiate pretty much anything! How cool is that?!

If you want to differentiate a polynomial, you just differentiate each term separately and then add them up together.

So - for example, if you want to differentiate $y = 2x + 4x^2$, I'd advise you to do this one step at a time (for now, once you get the hang of it, you can skip steps).

But for now, I want you to do something like this:



Please point out anything that you didn't understand in here.

Can you do these for me:

$y = 3x + 4$
$y = 3x^2 + 5x$

OK is the first one 3?
Is the second 6x+5?

OK is the first one 3?
Is the second 6x+5?

Both are correct.

Some harder ones if you want to try them :

$y=2x^2 - 3$

$y = 3x^2 - 2x + 7$

$y = 5x^4 - 2x^2 + 3$

Yes and yes.

1.) 4x
2.) 6x-2
3.) 20x^3-4x

1.) 4x
2.) 6x-2
3.) 20x^3-4x





Thanks I think I understand this now!

Can you tell me what qualification you are doing? E.g. IGCSE Edexcel?

Then we will know if you will need to practice harder questions or not.