# Differentiation

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Hey guys,

There something that I can't digest in this chapter: Differentiation

I understood the concept, but I just become confuse either for it symbol, calculation, or maybe I'm missing some concept here. Well, I'm actually talking about the calculation in a book I read.

Example 1

Given that $y = 3x^2 + 4x - 6$, find $\frac{dy}{dx}$ using the first principle

Solution:

Given that $y = 3x^2 + 4x - 6$ (1)

then, $y + \delta y = 3(x + \delta x)^2 + 4(x + \delta x) - 6$ (2)

(2) - (1): $\delta y = [3(x + \delta x)^2 + 4(x + \delta x) - 6] - 3x^2 - 4x + 6$

$\delta y = 3x^2 + 6x\delta x + 3(\delta x)^2 + 4x + 4\delta x - 6 - 3x^2 - 4x + 6$

$\delta y = 6x\delta x + 3(\delta x)^2 + 4\delta x$

Divide both sides by $\delta x$

$\frac{\delta y}{\delta x} = 6x + 3\delta x + 4$

$\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\delta y}{\delta x}$

$\frac{dy}{dx} = \lim_{\delta x \to 0} (6x + 3\delta x + 4)$

$\frac{dy}{dx} = 6x + 4$

But, when I give my try to solve one of the question...

Find the value of $\lim_{n \to -1} \frac{n^2 - 1}{n + 1}$

$y = (n^2 - 1)(n + 1)^-1$

$y + \delta y = [(n+\delta n)^2 - 1] [n + \delta n + 1] ^-1$

$y + \delta y = \frac{n^2 + 2n\delta n + \delta n^2 - 1}{n + \delta n + 1}$

$\delta y = \frac{n^2 + 2n\delta n + \delta n^2 - 1}{n + \delta n + 1} - \frac{n^2 - 1}{n + 1}$

well, I think I'm just not good in fraction or actually I don't understand why I'm coming to this step actually(probably just follow from the given example)

hope anyone can give a hand, thank you.

EDIT: btw, the answer is = -2

Edited by Vastor
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I'm not sure I understand what you're doing with the limit. Why are you trying to take the derivative of [imath](n^2 - 1)/(n+1)[/imath]?

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I don't see the relation between the two problems.

I'm not sure I understand what you're doing with the limit. Why are you trying to take the derivative of [imath](n^2 - 1)/(n+1)[/imath]?

I think it's an attempt at using L'Hopital's rule.

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I'm not sure I understand what you're doing with the limit. Why are you trying to take the derivative of [imath](n^2 - 1)/(n+1)[/imath]?

sincerely, I don't know what I'm doing with the limit. but finding a value for (something) / 0 is just impossible.

I don't see the relation between the two problems.

I think it's an attempt at using L'Hopital's rule.

ahah, probably that, I just don't know where to start to solve the equation.

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-

But, when I give my try to solve one of the question...

Find the value of $\lim_{n \to -1} \frac{n^2 - 1}{n + 1}$

$y = (n^2 - 1)(n + 1)^-1$

$y + \delta y = [(n+\delta n)^2 - 1] [n + \delta n + 1] ^-1$

$y + \delta y = \frac{n^2 + 2n\delta n + \delta n^2 - 1}{n + \delta n + 1}$

$\delta y = \frac{n^2 + 2n\delta n + \delta n^2 - 1}{n + \delta n + 1} - \frac{n^2 - 1}{n + 1}$

well, I think I'm just not good in fraction or actually I don't understand why I'm coming to this step actually(probably just follow from the given example)

hope anyone can give a hand, thank you.

EDIT: btw, the answer is = -2

This has nothing to do with taking a derivtive, by any method.

$\lim_{n \to -1} \frac{n^2 - 1}{n + 1}$ = $\lim_{n \to -1} \frac{(n - 1)(n+1)}{n + 1}=-2$

I think you need to go back and understand limits a bit better. Then and only then are you ready to understand derivatives.

You are concentrating on calculating and "finding the answer" when you need to concentrate on what the concepts mean.

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ahah, probably that, I just don't know where to start to solve the equation.

So, when you just plug in -1 to the original, it is undefined. This is a case where you can use L'Hopital's rule. Let f(x)=n2-1 and g(x)=n+1.

Now you can set $\lim_{x{\rightarrow}-1}{\frac{f(x)}{g(x)}}=\lim_{x{\rightarrow}-1}{\frac{f'(x)}{g'(x)}}$. -2 is indeed the correct answer.

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This has nothing to do with taking a derivtive, by any method.

$\lim_{n \to -1} \frac{n^2 - 1}{n + 1}$ = $\lim_{n \to -1} \frac{(n - 1)(n+1)}{n + 1}=-2$

I think you need to go back and understand limits a bit better. Then and only then are you ready to understand derivatives.

You are concentrating on calculating and "finding the answer" when you need to concentrate on what the concepts mean.

Heh. I completely forgot to factor. Yes, this method works as well.

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-

This has nothing to do with taking a derivtive, by any method.

$\lim_{n \to -1} \frac{n^2 - 1}{n + 1}$ = $\lim_{n \to -1} \frac{(n - 1)(n+1)}{n + 1}=-2$

I think you need to go back and understand limits a bit better. Then and only then are you ready to understand derivatives.

You are concentrating on calculating and "finding the answer" when you need to concentrate on what the concepts mean.

after read this post, you are convincing me to re-read the tutorial once again, until I see how you solve the question.

Expected! it's just a problem of mine dealing with denominator that have addition/subtraction symbol. How do you actually factor it? I mean, if me use that method, it should be a wild guess only.

About "my method", well, actually I'm just filling the current rules to show my trials, so that people can see where I'm stuck, I'm using derivative for the #2 post in the link above(The Tutorial of Differentiation) where it use some sort of calculation to remove its denominator so that it can move variable "h" to the limit, "0".

because of my frustration with the denominator and my un-complete understanding of the Differentiation itself, I give a try to apply the method, thus end up here.

P.S. I can solve the second question where $\lim_{n \to 0} \frac{n^2 - 2n}{n}$ , thnx for not having the silly (+/-) in its denominator

So, when you just plug in -1 to the original, it is undefined. This is a case where you can use L'Hopital's rule. Let f(x)=n2-1 and g(x)=n+1.

Now you can set . -2 is indeed the correct answer.

Can anyone teach / give a link on how to use L'Hopital's rule, interesting. I'm new to calculus so can't really grasp anything from wikipedia, wolfram alpha, etc...

thnx.

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How do you actually factor it? I mean, if me use that method, it should be a wild guess only.

n2-1 is a difference of perfect squares. a2-b2=(a+b)(a-b). You can do the multiplication on the right hand side yourself to see that it's true.

Edited by Vastor
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Hey guys,

How do you find the derivative from first principles for:

$y = \frac{4}{x} + 5x$, and

$y = \frac{3}{x^2}$

anyone have any tip on how to solve the denominator that causing me to get indeterminant?

such as for second question: problem with indeterminant and to find the derivative at the same time.

$y = \frac{3}{x^2}$

$y + \delta y = \frac{3}{(\delta x + x)^2}$

$(y + \delta y)(\delta x + x)^2 = 3$

$yx^2 + 2yx\delta x + y\delta x^2 + x^2\delta y + 2x\delta x\delta y + x^2\delta y = 3$'

then...?! =S

thank you.

anyone?

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To answer the second specifically (and using $h = \Delta x$ to help avoid careless errors), set it up as the following:

$\lim_{h \to 0}\frac{\frac{3}{(x + h)^2} - \frac{3}{x^2}}{h}$

and simplify. The first can be handled in a similar fashion.

Edited by John
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Try using $f'(x)=\lim_{h{\rightarrow}0}\frac{f(x+h)-f(x)}{h}$.

edit: which is what John just said

Edited by ydoaPs
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Hey guys,

57. Variables x and y are related by the equation $x + y = 10$. A third variable p is defined by $p = x^2y$. Find the values of x and y such that p is maximum.

what a question, I'm not even understand what p is, firstly, it said

$p = x^2y$ then, "p is maximum", which from what I understand $\frac{dp}{dx} < 0$ or $p = 0$

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Hey guys,

57. Variables x and y are related by the equation $x + y = 10$. A third variable p is defined by $p = x^2y$. Find the values of x and y such that p is maximum.

what a question, I'm not even understand what p is, firstly, it said

$p = x^2y$ then, "p is maximum", which from what I understand $\frac{dp}{dx} < 0$ or $p = 0$

What you want to do is get p entirely in terms of x. Then find p'(x)=0 that gives you all of the turning points. Those points that make p''(x)<0, the turning point will be a maximum. Where a turning point makes p''(x)>0, the turning point is a minimum.

Edited by ydoaPs
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Hey guys,

62. The figure above shows a rectangle ABCD. Given that AB = 2k cm and AD = k cm. E and F are points on BC and DC respectively where BE = x cm and FC = 2x cm.

(a) Show that the area of $\Delta AEF$ is $(k^2 - kx + x^2) cm^2$

$2k * k - [\frac{2k*x}{2} + \frac{k*(2k - 2x)}{2} + \frac{2x * (k - x)}{2}] = \Delta AEF$

$2k * k - [ k*x + k*(k - x) + x *(k - x) ] = \Delta AEF$

$2k * k - [ kx + k^2 - kx + kx - x^2 ] = \Delta AEF$

$2k^2 - kx - k^2 + x^2 = \Delta AEF$

$k^2 - kx + x^2 = \Delta AEF$

perfect, but wthell it has anything to do with differentiation?! are there any alternative way to calculate this?

(b) Express x in terms of k such that the area of $\Delta AEF$ is minimum.

similar question to the post before, still vague on how to pick which one should be "the first derivative of *unknown1* with respect to *unknown2* which written as $\frac{d*unknown1*}{d*unknown2*}$ or $f'(*unknown2*)$

© Find the minimum area of $\Delta AEF$ in terms of k.

and the difference with "(b)" question is?!!!

thank you for the one who keep helping me, here some cookie.

Edited by Vastor
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oh, forgot to clarify about my "real" confusion.

how does you actually apply differentiation for geometry?

I'm hardly see any relationship there =S

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for (b):-

let $\Delta AEF = y$

$k^2 - kx + x^2 = y$

"*Delta*AEF is minimum"

$\frac{d^2y}{dx^2} > 0$ (I do it right?)

$\frac{dy}{dx} = 0$

$\frac{dy}{dx} = 2x - k = 0$ (I don't know why I take this step *finding derivative*, but it works, whatsoever, I really need to understand how that happen!)

$2x = k$

$x = \frac{k}{2}$

Unlike I thought before, from this sentence "Express x in terms of k"

something like $\frac{dx}{dk}$

the question sound like "find y in terms of k", so...

$k^2 - kx + x^2 = y$

$k^2 - k(\frac{k}{2}) + (\frac{k}{2})^2 = y$

$\frac{4k^2 - 2k^2 + k^2}{4} = y$

$\frac{3k^2}{4} = y$

hope someone bother to answer my confusion now.

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oh, forgot to clarify about my "real" confusion.

how does you actually apply differentiation for geometry?

I'm hardly see any relationship there =S

Really? You might want to learn from a different source than the one you're using if they didn't teach you this before they taught you how to do differentiation.

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Really? You might want to learn from a different source than the one you're using if they didn't teach you this before they taught you how to do differentiation.

well, most of the question still can be answered by using simultaneous equation and some pure logic.

like this question:-

A piece of wire of length 120 cm is bent to form a shape as shown in the figure above. Show that the area enclosed by the wire, in cm^2, is given by $480x - 60x^2$. Hence, find the maximum area of the figure and state the corresponding value of x.

let $area-of-diagram = A = 480x - 60x^2$

because it's said "maximum area of the figure", so it should be

$\frac{d^2A}{dx^2} < 0$

$\frac{dA}{dx} = 0$

$\frac{dA}{dx} = 480 - 120x = 0$

$480 - 120x = 0$

$120x = 480$

$x = 4$

"because the above *x* is the maximum value for the function, so, the maximum area should be"

$480x - 60x^2 = 480(4) - 60(4)^2$

$= 1920 - 960 = 960cm^2$

So, I thought I'm on the right road...

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Hey guys,

this rule doesn't seems clear to me,

If $\delta x$ and $\delta y$ represent small changes in x and y respectively, then

$\frac{\delta y}{\delta x} \approx \frac{dy}{dx}$

$\delta y \approx \frac{dy}{dx}* \delta x$

and one of the example given its answer that its 'equal' rather than 'approximation', what varies between the two?

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Hey guys,

this rule doesn't seems clear to me,

If $\delta x$ and $\delta y$ represent small changes in x and y respectively, then

$\frac{\delta y}{\delta x} \approx \frac{dy}{dx}$

$\delta y \approx \frac{dy}{dx}* \delta x$

and one of the example given its answer that its 'equal' rather than 'approximation', what varies between the two?

$\frac{dy}{dx}=\lim_{\delta{x}\rightarrow{0}}\lim_{\delta{y}\rightarrow{0}}\frac{\delta y}{\delta x}$

It's another way of writing $f'(x)=\lim_{h\rightarrow{0}}\frac{f(x+h)-f(x)}{h}$ which is the definition of a derivative. I'm not sure you actually understand what is going on when you're doing calculus. It's all about the rate of change (or slope if you're looking at it graphically) of a function. This should be very intuitive, but your questions seem to indicate that you're just trying to force raw calculation without understanding what you're doing. I'd suggest slowing down and maybe using a different text.

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$\frac{dy}{dx}=\lim_{\delta{x}\rightarrow{0}}\lim_{\delta{y}\rightarrow{0}}\frac{\delta y}{\delta x}$

It's another way of writing $f'(x)=\lim_{h\rightarrow{0}}\frac{f(x+h)-f(x)}{h}$ which is the definition of a derivative. I'm not sure you actually understand what is going on when you're doing calculus. It's all about the rate of change (or slope if you're looking at it graphically) of a function. This should be very intuitive, but your questions seem to indicate that you're just trying to force raw calculation without understanding what you're doing.

Well, I'm just insecure with my time management because I spend too much time just to understand everything in math's subject ONLY, there 3 more science subject to study for the exam incoming on November. Thus, I ask every question that I found "time-consuming" to figure out by myself or the info not really available on net.

Anyway, because of your respond to some of my question by not replying anything a.k.a. "you should think for yourself" respond (but it worked ), I think I just can't help myself but force my memory to work and remember the raw calculation. I still get some intuitive by using the method though I still unable to "read" calculus as proficient as algebra or simpler topic.

I'd suggest slowing down and maybe using a different text.

when you said so, does Khan Academy video that cover differentiation topic do well?

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• 3 weeks later...

yDoaps,

from your words, I'm sure I'm NOT misunderstand a bit but instead, there some error in the use of symbol here about the rule that I'm talking about.

ok let's use this example:-

Find the approximate increase in the area of a square when its side increases from 5 cm to 5.01 cm.

Given Solution:

Let the length of the side of the square be x cm. Then area, $A = x^2$

$\frac{dA}{dx} = 2x$

When x increases from 5 cm to 5.01 cm,

$\delta x = 5.01 - 5 = 0.01$

Hence, the approximate increase in area is

$\delta A \approx \frac{dA}{dx} * \delta x$

*this is where it get tricky*

$\delta A = 2x(0.01) = 0.02x$

When x = 5 cm, $\delta A = 0.02(5) = 0.1 cm^2$

My(alternative) solution:

the change of A if x changes from 5 cm to 5.01 cm can be calculated as follows, consindering $A = x^2$

$A = 5^2 = 25$

$A = 5.01^2 = 25.1001$

$\delta A = 25.1001 - 25 = 1.001$

now, that's what we call equal, right?

instead of $\delta A = 0.1 cm^2$, why not $\delta A \approx 0.1 cm^2$. that's my real question is.

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