Would this method still work in an expanding universe?

[matt grime]

However, his formula uses positions computed from a formula that needs constant acceleration to work.

But that has absolutely nothing to do with why you lost marks. And the more you assert this the more marks I would be tempted to deduct because you are displaying less and less understanding of what was going on in each post.

Secondly, if you're asked for a definition on a maths class why do you think it is acceptable to interpret that asking for a non-rigorous description? I really wouldn't try your explanation with the lecturer; he will look at you in a very funny way.

Thirdly, reread your description, it actually does not make sense as a sentence in the English language. It isn't even grammatically correct, never mind mathematically correct. It asserts something "for all x that exist" which is nonsense, the existence of the x or otherwise is immaterial, it is an existence of a limit (at x) that is important, and your sentence is not structurally capable of supporting a claim that that is what you meant.

 

[tony873004]

Now you are mixing apples and oranges. The entire discussion on how to calculate average velocities! Calculating the time to get from one point to another is a totally different issue!

You're right. I got it mixed up there. I should have said:

"I can not use the teacher's method to accurately compute how far an object will drop over a given period of time."

instead of

"I can not use the teacher's method to accurately compute how long it would take for an object several thousands of miles high to drop to the ground."

But the point I'm making is the same. s(t)=1100-100t-16t² is a constant acceleration formula and it is where delta d comes from.

 

[nrqed]

Let me try to phrase it one more way:
\frac{\delta d}{\delta t} works fine under any bizarre acceleration in this form.

But once you replace \delta d with the formulas that generate \delta d you get
<br /> \frac{1100-100(3)-16(3)^2-(1100-100(1)-16(1)^2)}{3-1}<br />

and this formula does require constant acceleration.

This is where the miscommunication lies. Why do you say this works only if the acceleration is constant? Of course, one has to use the function s(t) provided, but it does not matter at all that the acceleration is constant here!

 

[nrqed]

You're right. I got it mixed up there. I should have said:

"I can not use the teacher's method to accurately compute how far an object will drop over a given period of time."

instead of

"I can not use the teacher's method to accurately compute how long it would take for an object several thousands of miles high to drop to the ground."

But the point I'm making is the same. s(t)=1100-100t-16t² is a constant acceleration formula and it is where delta d comes from.

No, it is not where \delta d comes from! The definition v_{ave} = {\Delta x \over \Delta t} does not care on whether the acceleration is constant.

I think you are saying that the numerical value of delta d for that particular problem comes from using s(t). Of course it does! But nowhere does the fact that the acceleration is constant is required!

 

[tony873004]

...if you're asked for a definition on a maths class why do you think it is acceptable to interpret that asking for a non-rigorous description? I really wouldn't try your explanation with the lecturer; he will look at you in a very funny way...

If someone asks me for the definition of an apple, I'd reply with something like "round red fruit". I wouldn't think of giving a math formula.

Nowhere in the book does it mention "mathamatical definition". Nowhere in our lecture notes does it mention "mathamatical definition". So that's why if I'm asked for a definition I try to describe in words what it is.

He'll look at me funny anyway. We both have dry senses-of-humor and joke around with each other a lot.

 

[AKG]

You are ABSOLUTELY WRONG! There are three formulas:

v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

v_{avg} = \frac{s&#039;(t_2) - s&#039;(t_1)}{2}

s(t) = -16t^2 - 100t + 1100

The first is the definition of average velocity, and since it is a definition, it works for ALL s. The second is an equation for average velocity that only works when s''(t) is constant. The third is just some particular equation of the position of some hypothetical object. For that particular object, it happens that s''(t) is constant, so its acceleration is constant.

He applied formula 1 to the s that appears in formula 3. s happens to be a constant acceleration function, but whether s were constant acceleration or not makes no difference on whether he is allowed to apply 1 to 3. You applied 2 to 3, but as you've demonstrated, you don't know why it's okay to do this. The fact that applying 1 to 3 means plugging in values to formula 3, and the fact that formula 3 happens to be a constant acceleration equation has NOTHING to do with whether or not his approach was the right way to find average velocity. He did not require s to be a constant acceleration formula for his solution to work. You did, i.e. the fact that acceleration happens to be constant has EVERYTHING to do with whether or not your approach was correct, and the fact that you made no indication that you were aware of this on your tests makes his mark fair, if not overly generous.

Consider s(t) = t³. I can compute the average velocity over the interval [0, 100], it's just (100³ - 0³)/(100 - 0) = 10000. See, this method works, and it works in situations where s is not constant acceleration. Your method would give 15000, which is wrong.

Your confusing what it means for a solution to require that s be constant acceleration, and a solution to a problem where s happens to be constant acceleration. Basically, you're saying that since his solution to the problem is a solution to a problem involving a function s, where s'' is constant, his solution requires s'' to be constant. His solution to a problem involving s requires him to use the function s, and s is a constant acceleration function, so in a stupid sense, it requires him to use a constant acceleration function. Of course, the solution to any problem involving s must use s. But he doesn't use the fact that s'' is constant. You do, but you don't indicate that you're aware of this, so you were lucky to only lose 1/2 a mark.

 

[tony873004]

But nowhere does the fact that the acceleration is constant is required!

Then why does the teacher's method get the wrong answer when choosing heights and times large enough for the gravity gradient to come into play in a sifnificant manner?

 

[tony873004]

...but as you've demonstrated, you don't know why it's okay to do this...

How have I demonstrated that I don't know why it's OK to do this? I know that it's ok to do this because for scales as small as 1100 feet, gravity can be considered constant. It's the same assumption that kinmatics chapters in physics books make all the time.

 

[nrqed]

Then why does the teacher's method get the wrong answer when choosing heights and times large enough for the gravity gradient to come into play in a sifnificant manner?

I am not following you... He gets the wrong answer for what? For the average velocity? It's because you use the wrong formula for s(t)!

all we are saying is that for a given s(t), his formula will always give the correct average velocity between two times, no matter what s(t) is.

If you give an incorrect s(t) and then says his formula does not give the correct answer, it's cheating
(and notice that your formula would not give the "correct" answer either)

Pat

 

[AKG]

I totally understand what you're saying. That formula works fine for any bizarre function, and mine doesn't. But the formula that generates the input for that formula (d₁and d₂ for delta d) assumes a constant acceleration.

So what? Yes, s happens to be a constant acceleration function. But his method would have worked even if s were not constant acceleration, since his method is a simple application of the definition.

I can not use the teacher's method to accurately compute how long it would take for an object several thousands of miles high to drop to the ground. Not because there's something wrong with \frac{\delta d}{\delta t}, but because there's something wrong with the way the inputs for that formula was generated.

No, this is ABSOLUTELY WRONG! His approach would have worked just fine. His approach is just applying the definition, and that can never be wrong. It might be wrong that a function s with s'' constant can never accurately model the free fall of a real-life physical body. But his method does not involve figuring out s. The problem is not to figure out s. The problem is, given an s, find the average velocity. The problem is:

Given this s, find the average acceleration

and his method works, and will always for for any given s. If the problem were:

If you drop an object from height h with mass m, derive a formula s(t) that gives its position at time t

and you give the answer s(t) = at² + bt + c, then you've made a mistake in the physics. It wouldn't be a math mistake, it would be a physics mistake. But this is neither here nor there, since THIS WAS NOT THE PROBLEM! The problem wasn't to determine whether s(t) is the correct function to model the free fall of the object, the problem was to assume s(t) is the correct function, since that's what's given, and then compute the average velocity.

 

[AKG]

How have I demonstrated that I don't know why it's OK to do this? I know that it's ok to do this because for scales as small as 1100 feet, gravity can be considered constant. It's the same assumption that kinmatics chapters in physics books make all the time.

Again, WRONG! It's okay to apply the formula because s'' is constant. That, and only that, is the correct reason. Suppose the problem posed was this:

An object is dropped from 100ft above the ground, and it falls freely. It's position is given by s(t) = t³ + e^t - cos(t) + 1/(t+1) + 99. Find the average velocity from t=0 to t=1.

It doesn't matter that s is in enormous contradiction with physical facts. Suppose the above question were given, how would you answer it? Would you refuse to do it because it is physically incorrect? Or would you realize that, if the position is GIVEN by a function s(t), then no matter how bizarre s is, the average velocity can be computed, and using his method, it will always be correctly calculated.

 

[tony873004]

I realize what you're saying AKG, but since his formula s(t) defined this as a constant acceleration problem, I felt free to use another method that works fine for constant acceleration.

Since this was the test following the derivative chapter, it made sense to me that he would want us to solve problems using derivatives, rather than using formulas memorized from physics class.

 

[nrqed]

How have I demonstrated that I don't know why it's OK to do this? I know that it's ok to do this because for scales as small as 1100 feet, gravity can be considered constant. It's the same assumption that kinmatics chapters in physics books make all the time.

Did you explain that in your test?

we are not saying you got the wrong answer, we are saying that the way you proceeded, you had to go through an extra step to mention that you could do it that way because you knew the acceleration was constant.
If you learn a technique that works in general and then you learn a special trick that works in only some situations, and you use the special trick in a test, you should explain why it is ok to use the special trick instead of the more general method.

 

[tony873004]

Again, WRONG! It's okay to apply the formula because s'' is constant. That, and only that, is the correct reason...

We haven't covered 2nd deriatives yet, so forgive me if I don't understand this. But is s'' the rate at which acceleration changes? If so, isn't saying "It's okay to apply the formula because s" is constant" the same thing as saying "It's okay to apply the formula because gravity (acceleration) can be considered constant"?

 

[nrqed]

We haven't covered 2nd deriatives yet, so forgive me if I don't understand this. But is s'' the rate at which acceleration changes? If so, isn't saying "It's okay to apply the formula because s" is constant" the same thing as saying "It's okay to apply the formula because gravity (acceleration) can be considered constant"?

s'' = the second derivative and it represents here the rate at which the velocity changes. And yes, saying s'' is constant is saying that the acceleration is constant (you should always say "the acceleration due to gravity is constant", not "gravity is constant").

Pat

 

[tony873004]

Thanks, Pat.

I appreciate the responses of everyone who has participated in this thread. I don't spend too much time in the math forums, and I'm learning a lot from you, especially in the terminology and methods (i.e. what is a definition, state that the special condition exists, allowing me to use a shortcut, etc.)

There was another "why was I marked wrong" question I asked in post #8 in this thread. It's a bit buried now :). Does anyone have any input on that question? I though I saw a response to it last night, but now I can't find it.

 

[Hurkyl]

I realize what you're saying AKG, but since his formula s(t) defined this as a constant acceleration problem, I felt free to use another method that works fine for constant acceleration.

The problem, as has been said before, is that you didn't say why you could use that method.

As an analogy, simply guessing "16" works fine for any problem whose answer is 16. But surely you wouldn't expect to get full marks if you tried that on the test.

 

[Curious3141]

Tony, I think you're protesting too much about why you feel you're right. There has, as yet, been no real sign that you appreciate why the method you used leads one to believe that you've got your concepts wrong (and this is a bigger error to a teacher than merely getting the answer wrong).

The reason I used this method is because this was a test on the chapter that introduced derivatives. I could have totally pictured him saying something like "I didn't want you to use the methods you learned in physics class. I'm testing you on derivatives. I wanted you to find the answer using derivatives."

Now, if you'd done it the PROPER way (based on definition) and your teacher had marked you wrong and given you this (frankly, cockamamey) explanation as to why, THEN we'd all be on your side and telling you your teacher is an idiot.

As it stands, for you to have *presumed* your teacher wants you to do something in a more difficult, indirect way that doesn't apply in general cases, then complaining when you got credit deducted, makes you seem more than a little weird in your thinking.

Think about it. You're GIVEN the formula for s(t). One way involves plugging in two time values, subtracting and diving by the time. It works NO MATTER WHAT form the s(t) takes, and doesn't entail ANY assumptions about the acceleration or anything else. This is the correct, and expected way to get the answer, and it happens to be EASIER.

The way you did it involves differentiating the expression, THEN substituting time values, adding and diving by two. It ONLY works because s(t) happens to be in a particular fortuitous form. No mention was made of the fact that you're aware that what you're doing only works because of that BIG COINCIDENCE (yes, it is a big coincidence, particle mechanics questions often have inconstant accelerations). AND it is a more difficult method. One would have to question why, if you really knew what you were doing, would you bother to take more steps to get to an answer when you could've used a more correct method right from the start.

Teachers, in general and with few exceptions, are not obsessive about things. And they know a lot more than their students usually give them credit for. If you present the problem correctly worked out as it SHOULD be worked out (and you can never go wrong by starting with the definition of something), then the teacher will give you full credit whether you've covered that in your syllabus or not. If the teacher doesn't, you have a legitimate gripe and argument to get a second evaluation. But using a less reliable method simply because you think you're somehow "expected" to do that is not the way to go.

Frankly, I would've given you half a mark. Not half the credit but literally 0.5 marks out of 2. In my humble opinion, that's all your working would've deserved. I think your teacher was generous.

 

[nrqed]

Calculate the average rate of change for the indicated value of x.
f(x)=x²-5; x=3

Since we were not prohibited from using the power rule, my answer was:

f'(x)=2x
f'(3)=2*3=6

My score: 1.25/2

What the teacher wanted is:

6+h.

My reasoning as to why he's wrong:

6+h would be the average rate of change for [x, x+h], but not for x.
x is a single point. It's like asking what's the batting average of a player who has 1 hit in 1 at bat. It's 1/1. Here too, with only a single point, the denominator of a fraction designed to compute an average must contain a number n=number of data points.

So the average rate of change for a single point should be the instantaneous value of that point.

Again, thanks to everyone for playing the part of the teacher in debating against me.

the question as stated here does not make sense. To calculate an average rate of change, on must always be provided with an interval. But I have the feeling you are not really giving the full question! his answer contains "h" and you don't even mention h in the question! Are you sure he did not asked the average rate of change between x and x +h or something like that?

You cannot replace average rate of change by a derivative. These are different things. (only the limit as the interval goes to zero do they become equal)

 

[Curious3141]

Calculate the average rate of change for the indicated value of x.
f(x)=x²-5; x=3

Since we were not prohibited from using the power rule, my answer was:

f'(x)=2x
f'(3)=2*3=6

My score: 1.25/2

What the teacher wanted is:

6+h.

My reasoning as to why he's wrong:

6+h would be the average rate of change for [x, x+h], but not for x.
x is a single point. It's like asking what's the batting average of a player who has 1 hit in 1 at bat. It's 1/1. Here too, with only a single point, the denominator of a fraction designed to compute an average must contain a number n=number of data points.

So the average rate of change for a single point should be the instantaneous value of that point.

Again, thanks to everyone for playing the part of the teacher in debating against me.

In this case, if that had been the question exactly as it was posed, and your teacher had expected an answer of (6 + h), yes, your teacher is wrong. I can't even begin to see where the h would suddenly come from.

Do post the whole question, you might actually be right in this instance !

 

[nrqed]

Calculate the average rate of change for the indicated value of x.
f(x)=x²-5; x=3

Since we were not prohibited from using the power rule, my answer was:

f'(x)=2x
f'(3)=2*3=6

My score: 1.25/2

What the teacher wanted is:

6+h.

My reasoning as to why he's wrong:

6+h would be the average rate of change for [x, x+h], but not for x.
x is a single point. It's like asking what's the batting average of a player who has 1 hit in 1 at bat. It's 1/1. Here too, with only a single point, the denominator of a fraction designed to compute an average must contain a number n=number of data points.

So the average rate of change for a single point should be the instantaneous value of that point.

Again, thanks to everyone for playing the part of the teacher in debating against me.

I get his answer if the question was "what is the average rate of change between x and x+h", where x=3. This is just (f(3+h) - f(3))/h.

If the question was not stated this way, then the question does not make sense.

But in any case, it would still be incorrect to use a derivative which gives the instantaneous rate of change.

I would say that you cannot argue that your answer is correct.

But if the question was written exactly liek this, then it should be dismissed entirely because it was not properly phrased.

 

[tony873004]

...Frankly, I would've given you half a mark. Not half the credit but literally 0.5 marks out of 2. In my humble opinion, that's all your working would've deserved. I think your teacher was generous.

You're not the only person to have said the teacher was generous. It's good to hear these opinions, because as a 42-year old returning college student, I still don't have the mindset of a teacher or student. I've got the mindset of a businessman who wants to argue that I've provided the work as dictated by the contract. Methods of project completion were not discussed. The job is finished. It works, now PAY.

It's tough getting used to the fact that school does not work this way.

 

[nrqed]

You're not the only person to have said the teacher was generous. It's good to hear these opinions, because as a 42-year old returning college student, I still don't have the mindset of a teacher or student. I've got the mindset of a businessman who wants to argue that I've provided the work as dictated by the contract. Methods of project completion were not discussed. The job is finished. It works, now PAY.

It's tough getting used to the fact that school does not work this way.

As a teacher I can give you the other side.

The difference with the business world is that here, it is not the *result* that matters the most. As a teacher I don't care primarily about whether the students get the final answer correct. I care first and foremost about whether they *understand*. Do they blindly plug the numbers in equations ordo they have some undersstanding of what they are doing, why they are doing this way, etc. This is especially important when dealing with special tricks that don't work in all problems like the one you used. Because I have to check that the student understands that it is a special trick and I want to check that he/she knows when to use it and when not to use it.

This is why it takes hours and hours to correct tests and assignments. If I cared only about the answer, I would be done in minutes. The aswer is right? Full credit. The answer is wrong? Zero! That would save me hundreds of hours of work every semester!

It does not work that way. If they get the right answer, I want to check they understand. If they get it wrong, I give partial credit if they seem to understand part of it.

Pat

 

[tony873004]

...But I have the feeling you are not really giving the full question!

Am I really that untrustable ?

I scanned it:
http://orbitsimulator.com/BA/test3.GIF

Are you a calculus teacher, nrqed?
btw. I'm still stuck on #2. I got the right answer because it was simple using the power rule, but we were not allowed to use the power rule here. The teacher noticed that my algebra and my answer didn't jive.

The teacher is very cool. If we make corrections on the missed problems, and turn them in, he adds a few percentage points to our grade. I stand to gain an additional 3.2%, so if anyone can point me in the right direction with #2, I'd appreciate it. I get stuck because I don't know if pulling /(x+h) up to the numerator making (x+h)^-1 was the right way to go. And if it is, I still don't know how to proceed.

I asked this last night. Galileo responded, but I don't quite follow what he's doing.
https://www.physicsforums.com/showthread.php?t=115242

 

[nrqed]

Am I really that untrustable ?

Not at all
I meant to say that I had a feeling you had forgotten to include a line of the question..

I scanned it:
http://orbitsimulator.com/BA/test3.GIF

Then, if you want my honest opinion, the question, as it stands, makes no sense. An interval must always be provided to calculate an average rate of change.

If I would have walked in for the test, sat down and read the question, I would not have been able to answer despite my PhD in theoretical physics. Because it is assuming something which is not standard (that he wanted the average rate between x and x+h).
I can only guess that he has done lots of example where average rate always implied between x and x+h and that he assumed that everybody was so used to that that he did not need to be clear in the question. But to be honest, it is not a sensical question as it stands.

I can see why, without an interval specified, you might ahve been tempted to use a derivative (even though this is not the correct answer for the question...but the question is incomplete so I can see why you could be tempted to do something else that what was actually asked).

Are you a calculus teacher, nrqed?

I teach physics.

btw. I'm still stuck on #2. I got the right answer because it was simple using the power rule, but we were not allowed to use the power rule here. The teacher noticed that my algebra and my answer didn't jive.

The teacher is very cool. If we make corrections on the missed problems, and turn them in, he adds a few percentage points to our grade. I stand to gain an additional 3.2%, so if anyone can point me in the right direction with #2, I'd appreciate it. I get stuck because I don't know if pulling /(x+h) up to the numerator making (x+h)^-1 was the right way to go. And if it is, I still don't know how to proceed.

I asked this last night. Galileo responded, but I don't quite follow what he's doing.
https://www.physicsforums.com/showthread.php?t=115242

I will have a look...

Pat

 

[nrqed]

Am I really that untrustable ?

I scanned it:
http://orbitsimulator.com/BA/test3.GIF

Are you a calculus teacher, nrqed?
btw. I'm still stuck on #2. I got the right answer because it was simple using the power rule, but we were not allowed to use the power rule here. The teacher noticed that my algebra and my answer didn't jive.

The teacher is very cool. If we make corrections on the missed problems, and turn them in, he adds a few percentage points to our grade. I stand to gain an additional 3.2%, so if anyone can point me in the right direction with #2, I'd appreciate it. I get stuck because I don't know if pulling /(x+h) up to the numerator making (x+h)^-1 was the right way to go. And if it is, I still don't know how to proceed.

I asked this last night. Galileo responded, but I don't quite follow what he's doing.
https://www.physicsforums.com/showthread.php?t=115242

well, writing 1/(x+h) as (x+h)^-1 is correct but does not help you very much. You have to put over a common denominator

{1 \over x+h} - {1 \over x} = {x -x - h \over x(x+h) }= {-h \over x(x+h)}

So far, no approximation. Then you must use the fact that h is very small compared to x to replace (x+h) by x. Then you get -h/x^2 and, after dividing by h, you get -1/x^2, as expected.

Pat

 

[tony873004]

Thanks, Pat. Physics is my favorite subject.

Am I on the right track here?
This might be a little more readable:
http://orbitsimulator.com/BA/question2.gif
<br /> \begin{array}{l}<br /> f(x) = x^2 + \frac{4}{x} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h)^2 + \frac{4}{{x + h}} - \left( {x^2 + \frac{4}{x}} \right)}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h)^2 + \frac{4}{{x + h}} - x^2 - \frac{4}{x}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{x^2 + 2hx + h^2 + \frac{4}{{x + h}} - x^2 - \frac{4}{x}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left( {x^2 + hx} \right)x^2 }}{{x^2 + hx}} + \frac{{\left( {x^2 + hx} \right)2hx}}{{x^2 + hx}} + \frac{{\left( {x^2 + hx} \right)h^2 }}{{x^2 + hx}} + \frac{{4x}}{{x^2 + hx}} - \frac{{\left( {x^2 + hx} \right)x^2 }}{{x^2 + hx}} - \frac{{4\left( {x + h} \right)}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x^4 + hx^3 }}{{x^2 + hx}} + \frac{{2hx^3 + 2h^2 x^4 }}{{x^2 + hx}} + \frac{{x^2 h^2 + h^3 x}}{{x^2 + hx}} + \frac{{4x}}{{x^2 + hx}} - \frac{{x^4 + hx^3 }}{{x^2 + hx}} - \frac{{4x + 4h}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x^4 + hx^3 + 2hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x + 4x - x^4 - hx^3 - 4x - 4h}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> \end{array}<br />

<br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{x\left( {x + h} \right)}}}}{h}<br />
If this is correct so far, I'm stuck here.

 

[nrqed]

Thanks, Pat. Physics is my favorite subject.

Am I on the right track here?
<br /> \begin{array}{l}<br /> f(x) = x^2 + \frac{4}{x} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h)^2 + \frac{4}{{x + h}} - \left( {x^2 + \frac{4}{x}} \right)}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h)^2 + \frac{4}{{x + h}} - x^2 - \frac{4}{x}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{x^2 + 2hx + h^2 + \frac{4}{{x + h}} - x^2 - \frac{4}{x}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left( {x^2 + hx} \right)x^2 }}{{x^2 + hx}} + \frac{{\left( {x^2 + hx} \right)2hx}}{{x^2 + hx}} + \frac{{\left( {x^2 + hx} \right)h^2 }}{{x^2 + hx}} + \frac{{4x}}{{x^2 + hx}} - \frac{{\left( {x^2 + hx} \right)x^2 }}{{x^2 + hx}} - \frac{{4\left( {x + h} \right)}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x^4 + hx^3 }}{{x^2 + hx}} + \frac{{2hx^3 + 2h^2 x^4 }}{{x^2 + hx}} + \frac{{x^2 h^2 + h^3 x}}{{x^2 + hx}} + \frac{{4x}}{{x^2 + hx}} - \frac{{x^4 + hx^3 }}{{x^2 + hx}} - \frac{{4x + 4h}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x^4 + hx^3 + 2hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x + 4x - x^4 - hx^3 - 4x - 4h}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> \end{array}<br />

<br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{x\left( {x + h} \right)}}}}{h}<br />
If this is correct so far, I'm stuck here.

you are making it more complicated than it needs to be!

You Can do the limits applied to each term separately. The x^2 bit does not require you to put over the denominator of the other function...Do it separately, it's the easy one. Do the 1/x separately and I gave you all the steps

Pat

 

[Hurkyl]

But as long as he got that far, we might as well nudge him the rest of the way!

tony: I think your problem is just that you haven't simplified your fraction. What is:

<br /> \frac{a}{b} \div c<br />

?

 

[Curious3141]

Thanks, Pat. Physics is my favorite subject.

Am I on the right track here?
This might be a little more readable:
http://orbitsimulator.com/BA/question2.gif
<br /> \begin{array}{l}<br /> f(x) = x^2 + \frac{4}{x} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h)^2 + \frac{4}{{x + h}} - \left( {x^2 + \frac{4}{x}} \right)}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h)^2 + \frac{4}{{x + h}} - x^2 - \frac{4}{x}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{x^2 + 2hx + h^2 + \frac{4}{{x + h}} - x^2 - \frac{4}{x}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{\left( {x^2 + hx} \right)x^2 }}{{x^2 + hx}} + \frac{{\left( {x^2 + hx} \right)2hx}}{{x^2 + hx}} + \frac{{\left( {x^2 + hx} \right)h^2 }}{{x^2 + hx}} + \frac{{4x}}{{x^2 + hx}} - \frac{{\left( {x^2 + hx} \right)x^2 }}{{x^2 + hx}} - \frac{{4\left( {x + h} \right)}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x^4 + hx^3 }}{{x^2 + hx}} + \frac{{2hx^3 + 2h^2 x^4 }}{{x^2 + hx}} + \frac{{x^2 h^2 + h^3 x}}{{x^2 + hx}} + \frac{{4x}}{{x^2 + hx}} - \frac{{x^4 + hx^3 }}{{x^2 + hx}} - \frac{{4x + 4h}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x^4 + hx^3 + 2hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x + 4x - x^4 - hx^3 - 4x - 4h}}{{x^2 + hx}}}}{h} \\ <br /> \\ <br /> \end{array}<br />

<br /> f&#039;(x) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{hx^3 + hx^3 + 2h^2 x^4 + x^2 h^2 + h^3 x - 4h}}{{x\left( {x + h} \right)}}}}{h}<br />
If this is correct so far, I'm stuck here.

See your other thread, I've solved it there. https://www.physicsforums.com/showthread.php?t=115242