I don't understand the logic behind differentiation Watch

99wattr89
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Chapter 8 in the C3 book is really highlighting this problem for me, hopefully you guys can help clear things up for me.

I understand how in a function, f(x), a change in x produces a corresponding change in y, and dy/dx gives the ratio of the changes, or gradient.

But in chapter 8 there's the product rule, whose derivation says that given dy/dx = u(dv/dx)+v(du/dx)+(du/dx)dv, you get dy/dx = u(dv/dx)+v(du/dx), because as dx approaches 0 so does (du/dx)dv. But I don't understand this. As dx approaches 0, all the terms in the expression approach 0 too, so how can that one alone be eliminated?
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nuodai
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The idea is that when we differentiate a function we look at the gradient of the cord between two points on the curve as the two points become closer together.

The gradient of a cord between (x_0, f(x_0)) and (x_0+h, f(x_0+h)) would be \dfrac{f(x_0+h)-f(x_0)}{(x_0+h) - x_0} = \dfrac{f(x_0+h)-f(x_0)}{h}. As we let h tend to zero, we get \displaystyle \lim_{h \to 0} \dfrac{f(x_0+h)-f(x_0)}{h}, and this is exactly how we define \dfrac{dy}{dx} (evaluated at x=x_0, where y=f(x)). Clearly, as h tends to zero, both the top and the bottom of the fraction tend to zero, which is why we don't get silly answers like infinity all the time.

Now, if y=f(x) = u(x)v(x), then plugging into our formula above, we get
\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x)}{h}

But notice that u(x)v(x+h) - u(x)v(x+h) = 0, and so we can re-write this as
\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x+h)}{h} + \lim_{h \to 0} \dfrac{u(x)v(x+h) - u(x)v(x)}{h}

We can factor out \lim_{h\to 0} u(x+h) (which is equal to u(x)) in the first part and v(x) in the second part to get
\dfrac{dy}{dx} = \displaystyle u(x) \left[ \lim_{h \to 0} \dfrac{v(x+h)-v(x)}{h} \right] + v(x) \left[ \lim_{h \to 0} \dfrac{u(x+h)-u(x)}{h} \right]

But appealing to our definition above, we find that this is simply
\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}

...which is the product rule.

In all the above cases, the top and bottom of the fractions both tend to zero as h tends to zero, since [at A-level] the functions are continuous, so when h \to 0, we have that x+h \to x and hence u(x+h)-u(x) \to u(x)-u(x) = 0, and so on.
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99wattr89
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(Original post by nuodai)
The idea is that when we differentiate a function we look at the gradient of the cord between two points on the curve as the two points become closer together.

The gradient of a cord between (x_0, f(x_0)) and (x_0+h, f(x_0+h)) would be \dfrac{f(x_0+h)-f(x_0)}{(x_0+h) - x_0} = \dfrac{f(x_0+h)-f(x_0)}{h}. As we let h tend to zero, we get \displaystyle \lim_{h \to 0} \dfrac{f(x_0+h)-f(x_0)}{h}, and this is exactly how we define \dfrac{dy}{dx} (evaluated at x=x_0, where y=f(x)). Clearly, as h tends to zero, both the top and the bottom of the fraction tend to zero, which is why we don't get silly answers like infinity all the time.

Now, if y=f(x) = u(x)v(x), then plugging into our formula above, we get
\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x)}{h}

But notice that u(x)v(x+h) - u(x)v(x+h) = 0, and so we can re-write this as
\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x+h)}{h} + \lim_{h \to 0} \dfrac{u(x)v(x+h) - u(x)v(x)}{h}

We can factor out \lim_{h\to 0} u(x+h) (which is equal to u(x)) in the first part and v(x) in the second part to get
\dfrac{dy}{dx} = \displaystyle u(x) \left[ \lim_{h \to 0} \dfrac{v(x+h)-v(x)}{h} \right] + v(x) \left[ \lim_{h \to 0} \dfrac{u(x+h)-u(x)}{h} \right]

But appealing to our definition above, we find that this is simply
\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}

...which is the product rule.

In all the above cases, the top and bottom of the fractions both tend to zero as h tends to zero, since [at A-level] the functions are continuous, so when h \to 0, we have that x+h \to x and hence u(x+h)-u(x) \to u(x)-u(x) = 0, and so on.
I don't think I understand how this works, I'm really confused.

h has to be 0 for u(x) to equal u(x+h) and v(x) to equal v(x+h), but if so, then the whole expression is equal to zero on both sides.

dy/dx means (change in y)/(change in x), but if the change in x is 0 so is the change in y, so doesn't that mean the gradient is just 0?

Have I failed to understand what 'lim' means?
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nuodai
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(Original post by 99wattr89)
I don't think I understand how this works, I'm really confused.

h has to be 0 for u(x) to equal u(x+h) and v(x) to equal v(x+h), but if so, then the whole expression is equal to zero on both sides.

dy/dx means (change in y)/(change in x), but if the change in x is 0 so is the change in y, so doesn't that mean the gradient is just 0?

Have I failed to understand what 'lim' means?
Perhaps. Normally when the numerator of a fraction is zero then the fraction itself is zero. However, if you have a fraction whose numerator and denominator both tend to zero, then the fraction might take a completely different value.

It might be clearer with a different example. Say we have y=f(x)=x^2. Then

\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{(x+h)^2 - x^2}{h}

 = \displaystyle \lim_{h \to 0} \dfrac{x^2+2hx+h^2-x^2}{h}

 = \displaystyle \lim_{h \to 0} \left[ 2x + h \right]

 = 2x

But notice that in the fraction \dfrac{(x+h)^2-x^2}{h}, both the top and bottom of the fraction tend to zero as we move h closer to zero.
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99wattr89
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(Original post by nuodai)
Perhaps. Normally when the numerator of a fraction is zero then the fraction itself is zero. However, if you have a fraction whose numerator and denominator both tend to zero, then the fraction might take a completely different value.

It might be clearer with a different example. Say we have y=f(x)=x^2. Then

\dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{(x+h)^2 - x^2}{h}

 = \displaystyle \lim_{h \to 0} \dfrac{x^2+2hx+h^2-x^2}{h}

 = \displaystyle \lim_{h \to 0} \left[ 2x + h \right]

 = 2x

But notice that in the fraction \dfrac{(x+h)^2-x^2}{h}, both the top and bottom of the fraction tend to zero as we move h closer to zero.
This is so frustrating, I really don't understand.
I follow that the equation you present equals 2x+h, but then the next step I just don't understand. For 2x+h to be 2x that means h=0, but if h reaches 0 then the whole equation beocomes 0, not 2x.
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innerhollow
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(Original post by 99wattr89)
I don't think I understand how this works, I'm really confused.

h has to be 0 for u(x) to equal u(x+h) and v(x) to equal v(x+h), but if so, then the whole expression is equal to zero on both sides.

dy/dx means (change in y)/(change in x), but if the change in x is 0 so is the change in y, so doesn't that mean the gradient is just 0?

Have I failed to understand what 'lim' means?
I'm much less qualified than nuodai to talk about this... but I think I know where you're coming from!

Your mistake is thinking that \frac{0}{0} = 0. This is WRONG!!! Zero divided by zero does NOT equal zero!

AFAIK, you need limits to evaluate \frac{0}{0}. For example the limit of y = \frac{sinx}{x} as x tends towards 0 (and importantly sinx also tends towards 0 as you know that sin(0) = 0) is actually 1
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nuodai
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(Original post by 99wattr89)
This is so frustrating, I really don't understand.
I follow that the equation you present equals 2x+h, but then the next step I just don't understand. For 2x+h to be 2x that means h=0, but if h reaches 0 then the whole equation beocomes 0, not 2x.
How?

Let's say that x=3, then 2x+h = 6+h. How can the limit of 6+h, as h tends to zero, be zero? x doesn't depend on h at all, so as h varies, x doesn't change.
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99wattr89
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(Original post by nuodai)
How?

Let's say that x=3, then 2x+h = 6+h. How can the limit of 6+h, as h tends to zero, be zero? x doesn't depend on h at all, so as h varies, x doesn't change.
Well, the 2x was derived from (2hx+h^2)/h, so if h=0 then that would be (2*0*x+0*0)/0.

I do agree that if you have the result of 2x+h and h is 0, then you get 2x, but my confusion is that if h is 0 then you don't get 2x in the first place.

(Original post by innerhollow)
I'm much less qualified than nuodai to talk about this... but I think I know where you're coming from!

Your mistake is thinking that \frac{0}{0} = 0. This is WRONG!!! Zero divided by zero does NOT equal zero!

AFAIK, you need limits to evaluate \frac{0}{0}. For example the limit of y = \frac{sinx}{x} as x tends towards 0 (and importantly sinx also tends towards 0 as you know that sin(0) = 0) is actually 1
Thank you for pointing that out, I was indeed thinknig that 0/0=0. As I understand it, it actually equals both 0 and infinity.

But IO don't understand how from (2*0*x+0*0)/0 you can get 2x.
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nuodai
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(Original post by 99wattr89)
Well, the 2x was derived from (2hx+h^2)/h, so if h=0 then that would be (2*0*x+0*0)/0.

I do agree that if you have the result of 2x+h and h is 0, then you get 2x, but my confusion is that if h is 0 then you don't get 2x in the first place.
Ah, I see!

Well, \dfrac{2hx+h^2}{h} = 2x+h -- the h on the bottom cancels with the terms on the top. We're allowed to simplify before we take limits; for example it's perfectly valid to say \lim_{h \to 0} \frac{h}{h} = 1.
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nuodai
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(Original post by 99wattr89)
Thank you for pointing that out, I was indeed thinknig that 0/0=0. As I understand it, it actually equals both 0 and infinity.
\dfrac{0}{0} is meaningless, and has no mathematical value. It doesn't equal zero, 1, infinity, 6, 2, pi or anything; it's just meaningless. However, we can look at the limits of fractions whose numerators and denominators both tend to zero, and this is what we do here.

If you were to give \dfrac{0}{0} a value then it wouldn't be clear what value it should take. For example, as x \to 0, both x and ax (where a is a number) tend to zero, but \dfrac{ax}{x} = a, and so \lim_{x \to 0} \dfrac{ax}{x} = a. However, taking the limits of the top and the bottom of the fraction separately (which we're not allowed to do) would give \lim_{x \to 0} \dfrac{ax}{x} = \dfrac{0}{0}. Is it then valid to say \dfrac{0}{0} = a? Well, no it isn't, which is why we have to consider limits case-by-case.
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LazyWorseThanInfidel
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expert explanation http://www.youtube.com/watch?v=rAof9Ld5sOg
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(Original post by nuodai)
Ah, I see!

Well, \dfrac{2hx+h^2}{h} = 2x+h -- the h on the bottom cancels with the terms on the top. We're allowed to simplify before we take limits; for example it's perfectly valid to say \lim_{h \to 0} \frac{h}{h} = 1.
At least now I understand why I don't understand! xD
Thank you so much for sticking with me to explain.

So, it seems that what I don't understand is how we can change h's value after simplification. It has to be =/= 0 for the simplification, but then becomes 0 after?

Also, in the derivation in your first post, h seems to have to be 0 for the simplification itself to work.
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Farhan.Hanif93
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(Original post by 99wattr89)
Well, the 2x was derived from (2hx+h^2)/h, so if h=0 then that would be (2*0*x+0*0)/0.

I do agree that if you have the result of 2x+h and h is 0, then you get 2x, but my confusion is that if h is 0 then you don't get 2x in the first place.



Thank you for pointing that out, I was indeed thinknig that 0/0=0. As I understand it, it actually equals both 0 and infinity.

But IO don't understand how from (2*0*x+0*0)/0 you can get 2x.
I think what you're not getting is that h is NOT zero at this stage i.e. we're not applying the limit yet but at a later stage when it makes more sense, h will tend to zero. The aim is to simplify it first. Also it's not the product rule which you're not understanding, it's differentiation from first principles. Google it and read up on it. Nuodai's example should have also been sufficient.
(at least, that's how I think of it. I may be wrong)
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innerhollow
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(Original post by 99wattr89)
Thank you for pointing that out, I was indeed thinknig that 0/0=0. As I understand it, it actually equals both 0 and infinity.
According to this page (read the comment by Richard Barrans), zero/zero is undefined and can take the value of any real number.
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nuodai
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(Original post by 99wattr89)
At least now I understand why I don't understand! xD
Thank you so much for sticking with me to explain.

So, it seems that what I don't understand is how we can change h's value after simplification. It has to be =/= 0 for the simplification, but then becomes 0 after?

Also, in the derivation in your first post, h seems to have to be 0 for the simplification itself to work.
I'm trying to find a way to explain limits without them being confusing, but if you've just done C3 (or even C4) then it's hard to put it in terms you'll understand off the bat... this stuff isn't normally covered until first year of uni.

In cases like these, letting h tend to zero is like looking at the limit of a sequence or a series. Say for example we have a geometric series 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots with initial term 1 and common ratio \frac{1}{2}.

The formula S_{\infty} = \dfrac{a}{1-r} reveals that
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \dfrac{1}{1-1/2} = \dfrac{1}{1/2} \boxed{= 2}

This means that if we add an infinite number of these terms to each other, we end up with 2. However, it's obviously physically impossible to add together an infinite number of terms; and this is why we say that 2 is the 'limit' of the series. When we deal with things we can't physically do (e.g. divide by zero, add infinite things together, and so on), we have to appeal to limits.

When we take a limit when adding an infinite number of things together, we look at what happens when we add a finite number of things together, as the finite number grows. So with the geometric progression above, we look at the nth partial sum S_n = 1 + \frac{1}{2} + \cdots + \left(\frac{1}{2}\right)^{n-1}. Now we can derive that
S_n = \frac{1-(1/2)^n}{1-1/2}
...and to find the infinite limit, we see what happens as n grows; but notice that all this time n is still finite. We end up with the above boxed result when we see its behaviour as it tends to infinity. The key word here is "tends to".

We're doing the same thing with h here. We can't let it actually be zero, but we can look at what happens when it's close to zero. In particular, when it's arbitrarily close to zero. I won't go too far into what we mean by "arbitrarily close to", because that's a whole different ball game, but essentially this is the reason why we can treat h as if it's not zero before we take the limit.
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Mr M
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Which Awarding Body are you with OP? From memory, only WJEC requires you to have any sort of understanding of this at this stage in your education so you might be battling with it unnecessarily (although it is quite illuminating to see how it works).
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(Original post by 99wattr89)
At least now I understand why I don't understand! xD
Thank you so much for sticking with me to explain.

So, it seems that what I don't understand is how we can change h's value after simplification. It has to be =/= 0 for the simplification, but then becomes 0 after?

Also, in the derivation in your first post, h seems to have to be 0 for the simplification itself to work.

I am not sure whether this is going to help you but h is never 0. h gets closer and closer to zero but it is never 0.
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I wish I was just half as amazing as nuodai :love:
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The limit as h goes to zero is NOT the same thing as "say h=0". It is "suppose we made h smaller and smaller. What would the expression be to if h got so close to zero that it was irrelevant? Then we could cancel such and such in the expressions".


(Original post by Mr M)
Which Awarding Body are you with OP? From memory, only WJEC requires you to have any sort of understanding of this at this stage in your education so you might be battling with it unnecessarily (although it is quite illuminating to see how it works).
OCR MEI have asked questions on the understanding of differentiation as well. C2 June 2009.

Understanding it is still a good idea as it makes further topics easier to learn and they make more sense.

Differentiation as a gradient probably isn't the best way of thinking about it - I think of it as: a derivative how fast one variable changes compared to another, and differentiation is the process of finding a function to describe that rate of change from a function containing the relationship between the two variables.
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99wattr89
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Wow, a lot of posts, I hope I haven't missed anyone, thank you all for helping.

(Original post by Farhan.Hanif93)
I think what you're not getting is that h is NOT zero at this stage i.e. we're not applying the limit yet but at a later stage when it makes more sense, h will tend to zero. The aim is to simplify it first. Also it's not the product rule which you're not understanding, it's differentiation from first principles. Google it and read up on it. Nuodai's example should have also been sufficient.
(at least, that's how I think of it. I may be wrong)
Reviewing the first principles presented in that video is helping me clear my thoughts, I understand the concept of differentiating and how to use it, but wasn't clear on some of the finer points of the underlying method.

(Original post by innerhollow)
According to this page (read the comment by Richard Barrans), zero/zero is undefined and can take the value of any real number.
(Original post by nuodai)
\dfrac{0}{0} is meaningless, and has no mathematical value. It doesn't equal zero, 1, infinity, 6, 2, pi or anything; it's just meaningless. However, we can look at the limits of fractions whose numerators and denominators both tend to zero, and this is what we do here.

If you were to give \dfrac{0}{0} a value then it wouldn't be clear what value it should take. For example, as x \to 0, both x and ax (where a is a number) tend to zero, but \dfrac{ax}{x} = a, and so \lim_{x \to 0} \dfrac{ax}{x} = a. However, taking the limits of the top and the bottom of the fraction separately (which we're not allowed to do) would give \lim_{x \to 0} \dfrac{ax}{x} = \dfrac{0}{0}. Is it then valid to say \dfrac{0}{0} = a? Well, no it isn't, which is why we have to consider limits case-by-case.
Ah, this explains a lot!

(Original post by nuodai)
I'm trying to find a way to explain limits without them being confusing, but if you've just done C3 (or even C4) then it's hard to put it in terms you'll understand off the bat... this stuff isn't normally covered until first year of uni.

In cases like these, letting h tend to zero is like looking at the limit of a sequence or a series. Say for example we have a geometric series 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots with initial term 1 and common ratio \frac{1}{2}.

The formula S_{\infty} = \dfrac{a}{1-r} reveals that
1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \dfrac{1}{1-1/2} = \dfrac{1}{1/2} \boxed{= 2}

This means that if we add an infinite number of these terms to each other, we end up with 2. However, it's obviously physically impossible to add together an infinite number of terms; and this is why we say that 2 is the 'limit' of the series. When we deal with things we can't physically do (e.g. divide by zero, add infinite things together, and so on), we have to appeal to limits.

When we take a limit when adding an infinite number of things together, we look at what happens when we add a finite number of things together, as the finite number grows. So with the geometric progression above, we look at the nth partial sum S_n = 1 + \frac{1}{2} + \cdots + \left(\frac{1}{2}\right)^{n-1}. Now we can derive that
S_n = \frac{1-(1/2)^n}{1-1/2}
...and to find the infinite limit, we see what happens as n grows; but notice that all this time n is still finite. We end up with the above boxed result when we see its behaviour as it tends to infinity. The key word here is "tends to".

We're doing the same thing with h here. We can't let it actually be zero, but we can look at what happens when it's close to zero. In particular, when it's arbitrarily close to zero. I won't go too far into what we mean by "arbitrarily close to", because that's a whole different ball game, but essentially this is the reason why we can treat h as if it's not zero before we take the limit.
I understand you here (thank god), and now I understand how the specific example you used is equal to 2x, that's the gradient of the tangent, where h=0, the point where the differentiation is most accurate. So you work out the formula first, and h is a variable, not a set value (which is how I was thinking of it) and then 'set' h to 0 afterwards to find the most accurate answer.

This takes me back to my intial problem, that in you first post your explanation uses the effect of h=0 at the end. I'm still trying to understand how this works, can you apply h=0 to some parts of the equation but not others? I'm talking about this step;

We can factor out \lim_{h\to 0} u(x+h) (which is equal to u(x)) in the first part and v(x) in the second part to get
\dfrac{dy}{dx} = \displaystyle u(x) \left[ \lim_{h \to 0} \dfrac{v(x+h)-v(x)}{h} \right] + v(x) \left[ \lim_{h \to 0} \dfrac{u(x+h)-u(x)}{h} \right]
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