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Differentiation

A curve has equation y = 2x3 + x2 - 8x + 3.
Find dy / dx
Hi

Can someone please guide me through answering the above question step by step?

Thanks.
(edited 7 years ago)

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Reply 1
Original post by _Xenon_
A curve has equation y = 2x3 + x2 - 8x + 3.
Find dy / dx
Hi

Can someone please guide me through answering the above question step by step?

Thanks.

We need to know what you know already.

Do you know how to find dydx\frac{dy}{dx} if y=2x3y=2x^3 ?
Original post by _Xenon_
A curve has equation y = 2x3 + x2 - 8x + 3.
Find dy / dx
Hi

Can someone please guide me through answering the above question step by step?

Thanks.


http://m.wikihow.com/Differentiate-Polynomials
Reply 3
Original post by notnek
We need to know what you know already.

Do you know how to find dydx\frac{dy}{dx} if y=2x3y=2x^3 ?


Is it the power times by 2x (because there's a number in front of it)?

So 3 times by 2x , I can't remember the rest
Reply 4
Original post by _Xenon_
Is it the power times by 2x (because there's a number in front of it)?

So 3 times by 2x , I can't remember the rest

3 times by 2x then subtract 1 from the exponent
Reply 5
Original post by notnek
We need to know what you know already.

Do you know how to find dydx\frac{dy}{dx} if y=2x3y=2x^3 ?


By the way, we have not been taught this in school due to time so I'm learning some of the content myself hence the question as I'd like someone to expalin how these questions are done. :-)
Reply 6
Original post by _Xenon_
Is it the power times by 2x (because there's a number in front of it)?

So 3 times by 2x , I can't remember the rest

You're nearly there. Next you need to subtract one from the power.

So the 3 becomes a 2 :

2x36x22x^3 \rightarrow 6x^2

Now do you have an idea how to differentiate y=2x3+x28x+3y = 2x^3 + x^2 - 8x + 3 ?

It looks like you need to over the basics of this topic by looking at your textbook / internet.
Reply 7
Original post by _Xenon_
By the way, we have not been taught this in school due to time so I'm learning some of the content myself hence the question as I'd like someone to expalin how these questions are done. :-)

Ignore the end of my last post - you need to go over the basics.

Can you tell me what you get if you differentiate these?

y=3x2y=3x^2

y=6x4y=6x^4

y=2x5y=2x^5
(edited 7 years ago)
Reply 8
Original post by notnek
Ignore the end of my last post - you need to go over the basics.

Can you tell me what you get if you differentiate these?

y=3x2y=3x^2

y=6x4y=6x^4

y=2x5y=2x^5


I times by the power and reduce the power by one, right?

1.) 6x
2.) 24x^3
3.) 10x^4
Reply 9
Original post by _Xenon_
I times by the power and reduce the power by one, right?

1.) 6x
2.) 24x^3
3.) 10x^4

Correct :smile:

Now see if you can try these - they're a bit different

1) y=5xy = 5x
2) y=3y = 3
Reply 10
Original post by notnek
Correct :smile:

Now see if you can try these - they're a bit different

1) y=5xy = 5x
2) y=3y = 3


OK I'm guessing these using what I already know...

1.) 5x = 5x^1 = 5x^0 = 1
2.) 3 = 3^1 = 3^0 = 1

Are they both 1? Not sure...
Reply 11
Original post by _Xenon_
OK I'm guessing these using what I already know...

1.) 5x = 5x^1 = 5x^0 = 1
2.) 3 = 3^1 = 3^0 = 1

Are they both 1? Not sure...

5x05x^0 is not equal to 1.

5x0=5×x0=...5x^0 = 5 \times x^0 = ...


For the second one 3=3x03 = 3x^0. Now try differentiating 3x03x^0


I have to go out now but hopefully someone else will help you.
Reply 12
Original post by _Xenon_
OK I'm guessing these using what I already know...

1.) 5x = 5x^1 = 5x^0 = 1


This is almost correct. Think about is, what is 5x^0.


2.) 3 = 3^1 = 3^0 = 1

Are they both 1? Not sure...


This one, not so much. Remember, if you're differentiating w.r.t. xx (i.e you're doing ddx)\frac{\mathrm{d}}{\mathrm{d}x}) then you only change the exponents of xx. 3 is not x.

So, we have, actually that: 3=3x03 = 3x^0. So what's ddx(3)=dd(3x0)\frac{\mathrm{d}}{\mathrm{d}x}(3) = \frac{\mathrm{d}}{\mathrm{d}}(3x^0)?

Hint:

Spoiler

Reply 13
Original post by notnek
5x05x^0 is not equal to 1.

5x0=5×x0=...5x^0 = 5 \times x^0 = ...


For the second one 3=3x03 = 3x^0. Now try differentiating 3x03x^0


I have to go out now but hopefully someone else will help you.


OK thanks I thought that because anything to the power of 0 was meant to be equal to 1. However, is just x^0 equal to 1? So is it 5 times by 1 which equals 5? Not sure...


Original post by Zacken
This is almost correct. Think about is, what is 5x^0.



This one, not so much. Remember, if you're differentiating w.r.t. xx (i.e you're doing ddx)\frac{\mathrm{d}}{\mathrm{d}x}) then you only change the exponents of xx. 3 is not x.

So, we have, actually that: 3=3x03 = 3x^0. So what's ddx(3)=dd(3x0)\frac{\mathrm{d}}{\mathrm{d}x}(3) = \frac{\mathrm{d}}{\mathrm{d}}(3x^0)?

Hint:

Spoiler



OK so 3=3x^0 so is that 3*1 which equals 3??!
Reply 14
Original post by _Xenon_
OK thanks I thought that because anything to the power of 0 was meant to be equal to 1. However, is just x^0 equal to 1? So is it 5 times by 1 which equals 5? Not sure...


5 times 1 which equals 5, yes.


OK so 3=3x^0 so is that 3*1 which equals 3??!


Not, quite. Remember - you bring the power down, so for example: 3x03x^0 becomes 3×0×x01=?3\times 0 \times x^{0-1} = ?
Original post by _Xenon_
OK I'm guessing these using what I already know...

1.) 5x = 5x^1 = 5x^0 = 1
2.) 3 = 3^1 = 3^0 = 1

Are they both 1? Not sure...


Just a quick point. From what you've written it looks like you're saying that 5x^1 = 5x^0, where you mean ddx(5x1)=5x0\dfrac{\mathrm{d}}{\mathrm{d}x}(5x^1) = 5x^0.
Reply 16
Okay, here are some rules that you should get used to for differentiating. What dydx\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:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\mathrm{d}}{\mathrm{d}x}(f(x) + g(x)) = \frac{\mathrm{d}}{\mathrm{d}x}(f(x)) + \frac{\mathrm{d}}{\mathrm{d}x} (g(x))end{equation*}



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 x1x^1, you get 1x0=11x^0 = 1. So if you were to differentiate 2x2x you can simply write 2x=x+x2x = x+x and then it follows that ddx(2x)=ddx(x)+ddx(x)=1+1=2\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

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{\mathrm{d}}{\mathrm{d}x}(\lambda x) = \lambda \frac{\mathrm{d}}{\mathrm{d}x}(x)\end{equation*}

.

That is, if you want to differentiate a constant multiple of xx, you can pull the multiple out, differentiate the x term and then put the constant back in. This makes total sense for differentiating 2x2x because we can do ddx(2x)=2ddx(x)=2(1)=2\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 xx raised to a power, you do this:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n-1} \end{equation*}



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! :eek3: How cool is that?! :woo:

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+4x2y = 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:

Unparseable latex formula:

\displaystyle[br]\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x}(2x + 4x^2) \\ &= \frac{\mathrm{d}}{\mathrm{d}x}(2x) + \frac{\mathrm{d}}{\mathrm{d}x}(4x^2) \\ & = 2\frac{\mathrm{d}}{\mathrm{d}x}(x) + 4\frac{\mathrm{d}}{\mathrm{d}x}(x^2) \\ & = 2\frac{\mathrm{d}}{\mathrm{d}x}(x^1) + 4\frac{\mathrm{d}}{\mathrm{d}x}(x^2) \\ & = 2(1 \times x^{1-1}) + 4(2 \times x^{2-1}) \\ &= 2(1 \times x^0) + 4(2 \times x) \\ &= 2(1) + 4(2x) \\ & = 2 + 8x \end{align*}



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

Can you do these for me:

y=3x+4y = 3x + 4
y=3x2+5xy = 3x^2 + 5x
Reply 17
Original post by Zacken
5 times 1 which equals 5, yes.




Not, quite. Remember - you bring the power down, so for example: 3x03x^0 becomes 3×0×x01=?3\times 0 \times x^{0-1} = ?


Hi is it 3*0= 0
So 0x^-1 (wow that's definitely wrong) please help me with this bit, I'm confused, thanks.
Reply 18
Original post by _Xenon_
Hi is it 3*0= 0
So 0x^-1 (wow that's definitely wrong) please help me with this bit, I'm confused, thanks.


Nopes, that's very correct.

The derivative of a constant is 0. :smile:

BTW, I've written up a post for you up there, read it, take a while to digest and understand what I've said, and then shower me with questions.
Reply 19
Original post by Zacken
Nopes, that's very correct.

The derivative of a constant is 0. :smile:

BTW, I've written up a post for you up there, read it, take a while to digest and understand what I've said, and then shower me with questions.


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

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