Differentiation Question - Help Needed to cancel out h

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lhh2003
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Question Use the definition of the derivative to show that for the function f(x) = x^2 , f ' (2) = 4

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I have reached the point where I have f ' (2) = lim h -> 0 4 + h .

How do I cancel out the h though ?
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RDKGames
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(Original post by lhh2003)
Question Use the definition of the derivative to show that for the function f(x) = x^2 , f ' (2) = 4

- - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - -- - -- - -- - - - - - -- - - - -

I have reached the point where I have f ' (2) = lim h -> 0 4 + h .

How do I cancel out the h though ?
h tends to zero so you're left with 4
Last edited by RDKGames; 10 months ago
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Notnek
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(Original post by lhh2003)
Question Use the definition of the derivative to show that for the function f(x) = x^2 , f ' (2) = 4

- - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - -- - -- - -- - - - - - -- - - - -

I have reached the point where I have f ' (2) = lim h -> 0 4 + h .

How do I cancel out the h though ?
I think you need to read a textbook or watch a video about limits because this is a very basic question if you understand limits.

You need to think about what happens to 4+h as h gets closer and closer to 0. E.g. when h=1 then 4+h=5, when h=0.5 then 4+h = 4.5, when h=0.1 then 4+h = 0.1. What is 4+h tending towards as h tends to 0?
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lhh2003
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(Original post by Sir Cumference)
I think you need to read a textbook or watch a video about limits because this is a very basic question if you understand limits.

You need to think about what happens to 4+h as h gets closer and closer to 0. E.g. when h=1 then 4+h=5, when h=0.5 then 4+h = 4.5, when h=0.1 then 4+h = 0.1. What is 4+h tending towards as h tends to 0?
I have watched a video on limits after reading your comment.

What I gathered was that the limit is always 0 because it gives the most accurate (the exact so to speak) value of the gradient of the tangent of a point on a curve.

What confuses me, however, is how the derivative formula works if h = 0 :

Let y = f(x)

Point A ( x0 , f(x0) )

Point B (x0 + h , f(x0 + h) )

Let x0 = 5 and h = 0 , with lim h--> 0

f ' (5) = ( f ( 5 + 0) - f( 5 ) ) / 0


^^^ Is impossible , as you can't divide by zero. Also, if you did use 0 as a limit, h = 0 would not give you two separate points to calculate the gradient of the line joining them. This is what is confusing me.

Thank you.
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Notnek
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(Original post by lhh2003)
I have watched a video on limits after reading your comment.

What I gathered was that the limit is always 0 because it gives the most accurate (the exact so to speak) value of the gradient of the tangent of a point on a curve.

What confuses me, however, is how the derivative formula works if h = 0 :

Let y = f(x)

Point A ( x0 , f(x0) )

Point B (x0 + h , f(x0 + h) )

Let x0 = 5 and h = 0 , with lim h--> 0

f ' (5) = ( f ( 5 + 0) - f( 5 ) ) / 0


^^^ Is impossible , as you can't divide by zero. Also, if you did use 0 as a limit, h = 0 would not give you two separate points to calculate the gradient of the line joining them. This is what is confusing me.

Thank you.
Good question. Say you have this expression

\displaystyle \frac{h^2+5h}{h}

And you consider the limit as h tends to 0. Remember, this doesn't just mean plug 0 into the expression because it would be undefined as you say. Instead you need to find the value that the expression tends towards as h tends to 0. So if you plug in 0.1 you get

\displaystyle \frac{0.1^2+5 \times 0.1}{0.1}

and if you cancel that you get 0.1+5 so you can see that even though it's not defined when h = 0, it's still defined for small values of h. And as h tends to 0, the expression clearly tends to 5. Often it's best to simplify the expression fully before considering the limit. Does this make sense?
Last edited by Notnek; 10 months ago
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lhh2003
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(Original post by Sir Cumference)
Good question. Say you have this expression

\displaystyle \frac{h^2+5h}{h}

And you consider the limit as h tends to 0. Remember, this doesn't just mean plug 0 into the expression because it would be undefined as you say. Instead you need to find the value that the expression tends towards as h tends to 0. So if you plug in 0.1 you get

\displaystyle \frac{0.1^2+5 \times 0.1}{0.1}

and if you cancel that you get 0.1+5 so you can see that even though it's not defined when h = 0, it's still defined for small values of h. And as h tends to 0, the expression clearly tends to 5. Often it's best to simplify the expression fully before considering the limit. Does this make sense?
I understand what you are saying.

To my understanding, the phrase 'tends towards' suggests that an output value of a function naturally moves towards a specific value based off of the input.

So, in the context of the original question, isn't the phrasing 'h tends towards 0' technically incorrect ? h is the independent variable, i.e we decide what h is. So surely we would say that f ' (2) tends towards 4, as h approaches the limit ?

Also in the context of the original question, it doesn't make sense to say that lim h --> 0 , therefore , f ' (2) = 4 . Surely an approximation symbol should be used instead ?
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Notnek
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(Original post by lhh2003)
I understand what you are saying.

To my understanding, the phrase 'tends towards' suggests that an output value of a function naturally moves towards a specific value based off of the input.

So, in the context of the original question, isn't the phrasing 'h tends towards 0' technically incorrect ? h is the independent variable, i.e we decide what h is. So surely we would say that f ' (2) tends towards 4, as h approaches the limit ?

Also in the context of the original question, it doesn't make sense to say that lim h --> 0 , therefore , f ' (2) = 4 . Surely an approximation symbol should be used instead ?
I think you may need to revisit the definition of a derivative. f'(2) is the gradient of the tangent to the curve f(x) at the point x = 2. For e.g. f(x)=x^2, this does have an exact value (as opposed to an approximate value) which is 4.

If you draw a line connecting coordinates (2, f(2)) and (2+h, f(2+h)) where h is some small value then this is a chord which has a variable gradient depending on the value of h. The key point is that as h gets closer to 0, the gradient of the chord gets closer and closer to the gradient of the tangent (i.e. the derivative) but never reaches it. So using the definition of the derivative you have

\displaystyle f'(2) = \lim_{h\rightarrow 0}{\frac{f(2+h) - f(2)}{h}}

If h is really close to 0 then you would use the "approximate to" symbol. But lim means the value that the expression is tending towards but will never reach it. So this is equal to f'(2) and not approximately equal to it. Make sense?
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lhh2003
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(Original post by Sir Cumference)
I think you may need to revisit the definition of a derivative. f'(2) is the gradient of the tangent to the curve f(x) at the point x = 2. For e.g. f(x)=x^2, this does have an exact value (as opposed to an approximate value) which is 4.

If you draw a line connecting coordinates (2, f(2)) and (2+h, f(2+h)) where h is some small value then this is a chord which has a variable gradient depending on the value of h. The key point is that as h gets closer to 0, the gradient of the chord gets closer and closer to the gradient of the tangent (i.e. the derivative) but never reaches it. So using the definition of the derivative you have

\displaystyle f'(2) = \lim_{h\rightarrow 0}{\frac{f(2+h) - f(2)}{h}}

If h is really close to 0 then you would use the "approximate to" symbol. But lim means the value that the expression is tending towards but will never reach it. So this is equal to f'(2) and not approximately equal to it. Make sense?
And how do you decide the value that the expression is tending towards ? Is this just the value of the expression when h = 0 ?
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RDKGames
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(Original post by lhh2003)
And how do you decide the value that the expression is tending towards ? Is this just the value of the expression when h = 0 ?
You simplify \dfrac{f(2+h) - f(2)}{h} as much as you can, and then observe each term and see what happens to it as h \to 0.

Example: Determine the derivative of f(x) = x^3 at the point x=1.

This is given by

\begin{aligned}f'(1) & = \lim_{h \to 0} \dfrac{f(1+h) - f(1)}{h} \\[0.3cm] & = \lim_{h \to 0} \dfrac{(1+h)^3 - (1)^3}{h} \\[0.3cm] & = \lim_{h \to 0} \dfrac{(1 + 3h + 3h^2 + h^3) - (1)}{h} \\[0.3cm] & =\lim_{h \to 0} \dfrac{3h + 3h^2 + h^3}{h} \\[0.3cm] & =\lim_{h \to 0} (3 + 3h + h^2) \end{aligned}

At this stage, there is nothing wrong with substituting in h=0 in this because this is an example of a 'nice' function where as h tends to zero, the result is the same as just plugging in h=0.

At A-Level I doubt you will ever be exposed to anything more (which in a way is a shame because it doesn't help you understand *why* we are interested in limits in the first place if we can just sub in h=0) although in Year 2 when you differentiate trig functions from first principles, it's a little bit more complicated than this which prevents you from plugging in h=0 even after simplying.

In more complicated cases, you would need to observe the behaviour of each term as h tends to zero.

In the example above, the limit can be split into three different ones;

\displaystyle \lim_{h \to 0} (3 + 3h + h^2) = \lim_{h \to 0} (3) + \lim_{h \to 0}(3h) + \lim_{h \to 0}(h^2)

For the first one, notice how the expression inside doesn't depend on h. So in the limit, nothing happens to it!

Hence \displaystyle \lim_{h \to 0} (3) = 3

For the second one, notice that if h \to 0, then 3h \to 0 as well. Thus we say \displaystyle \lim_{h \to 0}(3h) = 0.

For the last one, it's the same argument. Notice that if h \to 0 then its square h^2 definitely tends to zero as well. Thus \displaystyle \lim_{h \to 0}(h^2) = 0.

Hence the final answer;

 \displaystyle \lim_{h \to 0} (3 + 3h + h^2) = \lim_{h \to 0} (3) + \lim_{h \to 0}(3h) + \lim_{h \to 0}(h^2)  = 3 + 0 + 0 = 3

which is the correct answer for the problem.
Last edited by RDKGames; 10 months ago
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