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I'm halfway through Calculus 1 and I've learned nothing. I've had some serious problems with motivation and energy this semester and have literally put ZERO time into studying. It's starting to worry me because I'm afraid I can't play catch up now. But I thought starting was better than not starting. So I would like it if someone could please get me on the right track to understanding what I'm doing.

What is the point of calculus? (I understand it to be the mathematical study of change)

What is a limit?

What is a derivative?

What does it mean to differentiate?

What is an integral?

What way of thinking will help me understand calculus better?

Perhaps I can post some problems further on into this thread and get some help.

Thanks!

Edited by Tampitump
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Calculus is a broad field of mathematics. It originated from the efforts of Sir Isaac Newton and Gottfried Wilhelm Leibniz. Calculus basically studies how things in nature change and transform.

To understand calculus, you must have a good knowledge of basic arithmetic along with vectors and coordinate geometry.

Well, the basic concept behind limits is that there may be numbers that are very small but not zero. Or, numbers that are very very large but not infinity. In the language of calculus, we say that such numbers approaches the limit.

Now what's derivative? Basically derivative of any function is equivalent to the slope obtained when the function is plotted on graph.

Let f(x)=x^2+5, and y=f(x). Plotting a graph of x and y, you can easily find the slope as (y2-y1)/(x2-x1). In the context of calculus, we denote this as dy/dx=df(x)/dx. And that's the derivative of the function f(x).

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Some of these questions you should be able to find reasonable answers to via wikipedia...

But as to the point of calculus, well there are two at first seemingly different topics in calculus

i) Differentiation

ii) Integration

The first as you say deals with the instantaneous rates of change of functions (and similar objects). For example, a straight line can be described by y(x) = mx +c. Taking the derivative gives dy/dx = m. This is like the `velocity'. Taking the derivative (w.r.t. x) gives zero - the rate of change of the gradient or slope of a straight line is zero, thus it does not change.

There is then an inverse operation of differentiation - given up to an additive constant - this we call the antiderivative. Following the simple example, if I have a constant function m, then the antiderivative is y(x) = mx+c, but the c is arbitrary as there is no way to fix this without some further information.

On to integration, this is usually introduced as a calculational tool for working out the areas under a curve y(x) between x0 and x1 (say). This then can be generalised to find volumes and so on.

The loose idea is to cut up the area under the curve into thin strips and then add up all these strips. For a finite number of strips you get an approximation to this area. If you consider an infinite number of strips then you get the area - but to make sense of this you need limits. This gives us the definite integral as it is between two points.

There is also the indefinite integral, where no bounding points are given.

Now, the amazing thing is the fundamental theorem of calculus tells us that the antiderivative and the indefinite integral are the same thing. Thus differentiation and integration are tightly related.

Edited by ajb
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If you know the equation of position (x) for some object with time, f(x), the rate of change of position wrt time is speed. So the differential of f(x) is speed, f'(x). The rate of change of speed is acceleration, so the second differential of f(x), f''(x). You can even go backwards so the area under the line for acceleration is the speed so you can integrate f''(x) to get the speed, f'(x). And again back to position.

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Great suggestion - you really can picture what is going on.

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Thanks for reminding me ajb, I was going to say that as well

+1 Klaynos

Tampitump, your approach now should very much depend upon the sort of questions you are likely to be asked in your calculus exam.

Do they want you to show understanding of the underlying principles?

Do they want you to be able to extract derivatives and integrals according to particular rules and methods?

Do they expect you to apply the knowledge of calculus to anything (eg maxima and minima) ?

Anything else?

Whilst I am suggesting you apply your main focus to your exam requirements, to help us know where you are coming from can you tell us what you already know pre calculus.

Do you for instance know what a function is?

What notation are you using?

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not comprehensive, but other useful things from it are that you can differentiate an equation on any point on a curve to get the gradient at that point and you can integrate the equation of a curve or line between 2 points to get the area beneath it on a graph - both useful in the mechanics suggestion from above.

Take the velocity vs. time graph... the gradient or slope of the curve is the rate that the speed is increasing dv/dt (change in velocity with time) - so you can differentiate an equation at any given point on the graph of v vs t to find dv/dt which is the rate of change of velocity compared to time, which is acceleration. - very useful.

Edited by DrP
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have you tried this site ?

helping engineers learn mathematics

HELM - Helping Engineers Learn Mathematics - is a major 3-year curriculum development project to support the mathematical education of engineering students. The project is sponsored by the Higher Education Funding Council for England (HEFCE) through phase 4 of the Fund for the Development for Teaching and Learning (FDTL4)

http://www.personal.soton.ac.uk/jav/soton/HELM/helm_workbooks.html

http://www118.zippyshare.com/v/QvYr0f2E/file.html

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PS - I wouldn't worry about finding it hard to understand first few times you get taught it, it is quite normal as it can be hard to get your head around it. Wait until you try to do questions involving integration by parts and other complex integrals, lol.

It gets even harder still when the integrals are given in complex algebraic forms which you have to first simplify into some form you recognise so you can then use integration by parts or the correct method to solve for your equation.

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I appreciate all the answers. I haven't read them all yet, but I will soon.

I printed the midterm review and tallied all of the various types of problems that occur on there. I will post these, as well as some of the problems. Perhaps the smart people here can help me work through them.

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I appreciate all the answers. I haven't read them all yet, but I will soon.

I printed the midterm review and tallied all of the various types of problems that occur on there. I will post these, as well as some of the problems. Perhaps the smart people here can help me work through them.

Perhaps now would be a good time to start?

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So basically what I'm getting from it so far is that a limit is basically making a number so infinitely close to a point on a curve so that you can take a derivative from it and find the instantaneous slope of that point (or velocity in some cases)? Is this accurate?

Please word it better than I did if you can. I really need this to click so I can visualize it.

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So basically what I'm getting from it so far is that a limit is basically making a number so infinitely close to a point on a curve so that you can take a derivative from it and find the instantaneous slope of that point (or velocity in some cases)? Is this accurate?

Please word it better than I did if you can. I really need this to click so I can visualize it.

It would have been good if you had answered my questions from post#6.

Particularly in respect of knowing what a function is.

However this is the right idea, but a complicated way to start learning differentiation.

We regard x and y as variables. This is they can represent any value within a given range.

If we are given a value of x and some expression connecting x and y we can then calculate the value of y.

So the value of y depends on the given value of x; it can no longer represent any value because of the expression.

We say that y is the dependent variable.

And that x is the independent variable since it can still represent any value we are given.

We write this as y = f(x) or we can write the actual expression such as y = x2

I will use this simple expression as my example for differentiation. This might be the height of a hanging electricity cable, dangling between two poles.

So as the poles get further apart the cable hangs down further.

Now the whole purpose of calculus is to enable us to calculate desired results.

We are usually more interested in finding the dependent variable, y, knowing the independent one x.

The function y = x2 is also known as the primitive and the derivative is more properly called the derived function.

You have probably seen it written

$\frac{{dy}}{{dx}} = 2x$
Now I am going to write this expression in a slightly different way.
$\frac{{\Delta y}}{{\Delta x}} = 2x$
Where
$\Delta y = \left( {{y_2} - {y_1}} \right)$
and
$\Delta x = \left( {{x_2} - {x_1}} \right)$
We can rearrange the expression for the derived function thus
$\Delta y = 2x\Delta x$
So a change in y is approximately equal to a change in x multiplied by twice the value of x.
So if we start from x = 1 and consider changes to y, resulting from smaller and smaller changes to x, we find that the above expression becomes more and more accurate.
I have tabled this in the spreadsheet, starting at x = 1 and making smaller and smaller changes to delta x, from 1 down to to zero in 0.1 steps.
As delta x gets smaller (the purple column) the value of y calculated from the approximate formula (the yellow column) gets closer and closer to the true value (the grey (column) calculated from the 'primitive'.
So the error is 1 for a step size of 1 and 0.01 for a step size of .1
If you followed this motivation for why we bother with all this and you like it, then we can move on to the question of a limit, which it also demonstrates quite well.
Edited by studiot
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I apologize studiot. I'm just a little scatter-brained. You kind of have to bear with me on these matters. I'm sure you are used to conversing with some very sharp and knowledgeable people here on this forum, so it might be the case that I test your patience a little more than others when it comes to maths. I usually have to flounder for a while before I start to understand.

As I promised above, I have written down all of the problems on my test review, and here are all the problems arranged in their respective categories:

Differentiate:

y = 6e^x + 4/3√x (the last part is cube root of x)

g(u) = √2u + √3u

g(x) = 5e^x √x

F(y) = (1/y^2 – 3/y^4) (y + 5y^3)

y = x^3/1 - x^2

f(x) = ax + b/cx + d

y = b cos t + t^2 sin t (with respect to t)

y = 6 – sec x/tan x

f(x) = log10 (x^7 + 8)

f(x) + sin (x) ln (8x)

G(y) = ln (4y + 1)^3/√(y^2 ÷ 1)

Find the Derivative:

F(x) = (x^4 + 9x^2 -7)^8

y = cos (a^9 + x^9)

y = 9^(5 – x^2)

g(x) = cosh (ln(x))

y = sin^-1 (4x + 1)

Implicit Differentiation dy/dx:

4x^3 + x^2y + xy^3 = 2

y cos x = 3x^2 + 2y^2

Implicit Differentiation to find equation of tangent line to the curve:

5x^2 + xy + 5y^2 = 11, (1,1) (ellipse)

Find equation of tangent line to the curve:

y = 10x sin x, (π/2,5π)

y = x^4 + 5x^2 – x, (1.5)

Logarithmic Differentiation:

y = √(x – 5/x^6 +4)

y – (cos 8x)^x

Linearization L(x):

f(x) = x^4 + 6x^2, a = -1

f(x) = x^(2/3), a = 8

Find Numerical Value:

Cosh 3

Cosh (ln3)

Word Problems:

Each side of a square is increasing at a rate of 8 cm/s. At what rate is the area of the square increasing when the area of the square is 49 cm^2?

The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing when the diameter is 60mm?

A plane flying horizontally at an altitude of 1 mi and a speed of 410 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station. (round your answer to the nearest whole number.)

I understand you guys cannot work all of these problems or address them all. I was hoping perhaps we could tackle a few (or several of them), at least identifying the concepts that need to be studied. I also posted them all so you could get a feel for the trends in concepts that are being used here and can better direct on what I need to focus my studies on. I appreciate all the help I can get, and all the help so far!

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Thanks for reminding me ajb, I was going to say that as well

+1 Klaynos

Tampitump, your approach now should very much depend upon the sort of questions you are likely to be asked in your calculus exam.

Do they want you to show understanding of the underlying principles?

Do they want you to be able to extract derivatives and integrals according to particular rules and methods?

Do they expect you to apply the knowledge of calculus to anything (eg maxima and minima) ?

Anything else?

Whilst I am suggesting you apply your main focus to your exam requirements, to help us know where you are coming from can you tell us what you already know pre calculus.

Do you for instance know what a function is?

What notation are you using?

There you are then you have answered most of my questions and you now need a crash course in differentiation methods.

Integration is yet to be tackled.

What you have listed in post#14 is basically the algebraic manipulation of derivatives and a few examples of use.

Yes differentiation is basically algebraic manipulation.

There are not many rules, you could probably learn enough to pick up enough marks to scrape by since you obviously understand basic algebra.

(Not a path I'd recommend but it can be done and would be the start of your recovery).

Edited by studiot
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There you are then you have answered most of my questions and you now need a crash course in differentiation methods.

Integration is yet to be tackled.

What you have listed in post#14 is basically the algebraic manipulation of derivatives and a few examples of use.

Yes differentiation is basically algebraic manipulation.

There are not many rules, you could probably learn enough to pick up enough marks to scrape by since you obviously understand basic algebra.

(Not a path I'd recommend but it can be done and would be the start of your recovery).

Okay, thanks. I'm studying right now. I've done some studying today and it would appear that to solve most of these differentiation problems, I need to learn the various rules that are used to solve them (i.e. product rule, chain rule, power rule, quotient rule), and when to use these various methods. I think I've got the power rule down (i.e. the derivative of x^4 would be 4x^3), but there are some problems on here that seem a little trickier. I've also been told that there are shortcuts to differentiating which our professor has said we are about to get into in this course.

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Okay, thanks. I'm studying right now. I've done some studying today and it would appear that to solve most of these differentiation problems, I need to learn the various rules that are used to solve them (i.e. product rule, chain rule, power rule, quotient rule), and when to use these various methods. I think I've got the power rule down (i.e. the derivative of x^4 would be 4x^3), but there are some problems on here that seem a little trickier. I've also been told that there are shortcuts to differentiating which our professor has said we are about to get into in this course.

It is normal to have a table of standard derivatives available in exams, though it is quicker if you know them.

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It is normal to have a table of standard derivatives available in exams, though it is quicker if you know them.

Having the table doesn't guarantee I know how to use them. I'm not sure how to take the derivative of certain expressions. Like √(2u) or √(3u)

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√(x) is the same as x^(0.5). The first step in the ones you cannot do is to convert into that form then use the power rule.

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With that, I add that you should also learn the derivatives of trigonometric functions.

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Having the table doesn't guarantee I know how to use them. I'm not sure how to take the derivative of certain expressions. Like √(2u) or √(3u)

The table only contains the very simple, I'm sure you can cope with

$y = a{x^n}$
$\frac{{dy}}{{dx}} = na{x^{\left( {n - 1} \right)}}$
u and v are universally used to show an intermediate function of x.
This is also known as function of a function or the chain rule, but both are fancy names for substitution.
Watch
$y = \sqrt {3{x^2}}$
$Let\;u = 3{x^2}$
$y = \sqrt u = {u^{\frac{1}{2}}}$
to differentiate split the derivative into two parts
$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}*\frac{{du}}{{dx}}$
$\frac{{du}}{{dx}} = 6x$
$\frac{{dy}}{{du}} = \frac{d}{{du}}\left( {{u^{\frac{1}{2}}}} \right) = \frac{1}{2}{u^{\left( {\frac{1}{2} - 1} \right)}} = \frac{1}{2}{u^{ - \frac{1}{2}}} = \frac{1}{{2\sqrt u }} = \frac{1}{{2\sqrt {3{x^2}} }}$
$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}*\frac{{du}}{{dx}} = \frac{1}{{2\sqrt {3{x^2}} }}*6x = \frac{{3x}}{{\sqrt {3{x^2}} }} = \frac{3}{{\sqrt 3 }} = \sqrt 3$
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Ok, I think I've got the product, chain, quotient, and power rules down pat (at least their methodologies). I had some help from my school's online tutor service this morning and that was really helpful. Its pretty easy after you practice a few examples. There are some situations I have trouble with though. I will post some examples later. I have some work to get done for another class right now so I will check back in later.

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Careful.

You might get to like some of this stuff.

Edited by studiot
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Welp, I failed my calculus midterm..... MISERABLY! Semester ruined.

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single_question_with_answer

the text passage contains a personal statement about the outcome of a calculus midterm exam, which is a mathematical assessment. the user expresses disappointment in their performance, indicating they failed the exam and feel their semester is now negatively impacted. the question is implicit in the context, as the user is essentially asking for advice or support after this academic setback. the answer is provided through the context of the situation, which is the user's expression of their failure and the impact on their semester. however, the answer is not a direct solution to a mathematical problem, but rather a statement of the user's emotional and academic state.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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I'm halfway through Calculus 1 and I've learned nothing. I've had some serious problems with motivation and energy this semester and have literally put ZERO time into studying. It's starting to worry me because I'm afraid I can't play catch up now. But I thought starting was better than not starting. So I would like it if someone could please get me on the right track to understanding what I'm doing.

What is the point of calculus? (I understand it to be the mathematical study of change)

What is a limit?

What is a derivative?

What does it mean to differentiate?

What is an integral?

What way of thinking will help me understand calculus better?

Perhaps I can post some problems further on into this thread and get some help.

Thanks!

Edited by Tampitump
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Calculus is a broad field of mathematics. It originated from the efforts of Sir Isaac Newton and Gottfried Wilhelm Leibniz. Calculus basically studies how things in nature change and transform.

To understand calculus, you must have a good knowledge of basic arithmetic along with vectors and coordinate geometry.

Well, the basic concept behind limits is that there may be numbers that are very small but not zero. Or, numbers that are very very large but not infinity. In the language of calculus, we say that such numbers approaches the limit.

Now what's derivative? Basically derivative of any function is equivalent to the slope obtained when the function is plotted on graph.

Let f(x)=x^2+5, and y=f(x). Plotting a graph of x and y, you can easily find the slope as (y2-y1)/(x2-x1). In the context of calculus, we denote this as dy/dx=df(x)/dx. And that's the derivative of the function f(x).

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Some of these questions you should be able to find reasonable answers to via wikipedia...

But as to the point of calculus, well there are two at first seemingly different topics in calculus

i) Differentiation

ii) Integration

The first as you say deals with the instantaneous rates of change of functions (and similar objects). For example, a straight line can be described by y(x) = mx +c. Taking the derivative gives dy/dx = m. This is like the `velocity'. Taking the derivative (w.r.t. x) gives zero - the rate of change of the gradient or slope of a straight line is zero, thus it does not change.

There is then an inverse operation of differentiation - given up to an additive constant - this we call the antiderivative. Following the simple example, if I have a constant function m, then the antiderivative is y(x) = mx+c, but the c is arbitrary as there is no way to fix this without some further information.

On to integration, this is usually introduced as a calculational tool for working out the areas under a curve y(x) between x0 and x1 (say). This then can be generalised to find volumes and so on.

The loose idea is to cut up the area under the curve into thin strips and then add up all these strips. For a finite number of strips you get an approximation to this area. If you consider an infinite number of strips then you get the area - but to make sense of this you need limits. This gives us the definite integral as it is between two points.

There is also the indefinite integral, where no bounding points are given.

Now, the amazing thing is the fundamental theorem of calculus tells us that the antiderivative and the indefinite integral are the same thing. Thus differentiation and integration are tightly related.

Edited by ajb
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If you know the equation of position (x) for some object with time, f(x), the rate of change of position wrt time is speed. So the differential of f(x) is speed, f'(x). The rate of change of speed is acceleration, so the second differential of f(x), f''(x). You can even go backwards so the area under the line for acceleration is the speed so you can integrate f''(x) to get the speed, f'(x). And again back to position.

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Great suggestion - you really can picture what is going on.

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Thanks for reminding me ajb, I was going to say that as well

+1 Klaynos

Tampitump, your approach now should very much depend upon the sort of questions you are likely to be asked in your calculus exam.

Do they want you to show understanding of the underlying principles?

Do they want you to be able to extract derivatives and integrals according to particular rules and methods?

Do they expect you to apply the knowledge of calculus to anything (eg maxima and minima) ?

Anything else?

Whilst I am suggesting you apply your main focus to your exam requirements, to help us know where you are coming from can you tell us what you already know pre calculus.

Do you for instance know what a function is?

What notation are you using?

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not comprehensive, but other useful things from it are that you can differentiate an equation on any point on a curve to get the gradient at that point and you can integrate the equation of a curve or line between 2 points to get the area beneath it on a graph - both useful in the mechanics suggestion from above.

Take the velocity vs. time graph... the gradient or slope of the curve is the rate that the speed is increasing dv/dt (change in velocity with time) - so you can differentiate an equation at any given point on the graph of v vs t to find dv/dt which is the rate of change of velocity compared to time, which is acceleration. - very useful.

Edited by DrP
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have you tried this site ?

helping engineers learn mathematics

HELM - Helping Engineers Learn Mathematics - is a major 3-year curriculum development project to support the mathematical education of engineering students. The project is sponsored by the Higher Education Funding Council for England (HEFCE) through phase 4 of the Fund for the Development for Teaching and Learning (FDTL4)

http://www.personal.soton.ac.uk/jav/soton/HELM/helm_workbooks.html

http://www118.zippyshare.com/v/QvYr0f2E/file.html

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PS - I wouldn't worry about finding it hard to understand first few times you get taught it, it is quite normal as it can be hard to get your head around it. Wait until you try to do questions involving integration by parts and other complex integrals, lol.

It gets even harder still when the integrals are given in complex algebraic forms which you have to first simplify into some form you recognise so you can then use integration by parts or the correct method to solve for your equation.

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I appreciate all the answers. I haven't read them all yet, but I will soon.

I printed the midterm review and tallied all of the various types of problems that occur on there. I will post these, as well as some of the problems. Perhaps the smart people here can help me work through them.

##### Share on other sites

I appreciate all the answers. I haven't read them all yet, but I will soon.

I printed the midterm review and tallied all of the various types of problems that occur on there. I will post these, as well as some of the problems. Perhaps the smart people here can help me work through them.

Perhaps now would be a good time to start?

##### Share on other sites

So basically what I'm getting from it so far is that a limit is basically making a number so infinitely close to a point on a curve so that you can take a derivative from it and find the instantaneous slope of that point (or velocity in some cases)? Is this accurate?

Please word it better than I did if you can. I really need this to click so I can visualize it.

##### Share on other sites

So basically what I'm getting from it so far is that a limit is basically making a number so infinitely close to a point on a curve so that you can take a derivative from it and find the instantaneous slope of that point (or velocity in some cases)? Is this accurate?

Please word it better than I did if you can. I really need this to click so I can visualize it.

It would have been good if you had answered my questions from post#6.

Particularly in respect of knowing what a function is.

However this is the right idea, but a complicated way to start learning differentiation.

We regard x and y as variables. This is they can represent any value within a given range.

If we are given a value of x and some expression connecting x and y we can then calculate the value of y.

So the value of y depends on the given value of x; it can no longer represent any value because of the expression.

We say that y is the dependent variable.

And that x is the independent variable since it can still represent any value we are given.

We write this as y = f(x) or we can write the actual expression such as y = x2

I will use this simple expression as my example for differentiation. This might be the height of a hanging electricity cable, dangling between two poles.

So as the poles get further apart the cable hangs down further.

Now the whole purpose of calculus is to enable us to calculate desired results.

We are usually more interested in finding the dependent variable, y, knowing the independent one x.

The function y = x2 is also known as the primitive and the derivative is more properly called the derived function.

You have probably seen it written

$\frac{{dy}}{{dx}} = 2x$
Now I am going to write this expression in a slightly different way.
$\frac{{\Delta y}}{{\Delta x}} = 2x$
Where
$\Delta y = \left( {{y_2} - {y_1}} \right)$
and
$\Delta x = \left( {{x_2} - {x_1}} \right)$
We can rearrange the expression for the derived function thus
$\Delta y = 2x\Delta x$
So a change in y is approximately equal to a change in x multiplied by twice the value of x.
So if we start from x = 1 and consider changes to y, resulting from smaller and smaller changes to x, we find that the above expression becomes more and more accurate.
I have tabled this in the spreadsheet, starting at x = 1 and making smaller and smaller changes to delta x, from 1 down to to zero in 0.1 steps.
As delta x gets smaller (the purple column) the value of y calculated from the approximate formula (the yellow column) gets closer and closer to the true value (the grey (column) calculated from the 'primitive'.
So the error is 1 for a step size of 1 and 0.01 for a step size of .1
If you followed this motivation for why we bother with all this and you like it, then we can move on to the question of a limit, which it also demonstrates quite well.
Edited by studiot
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I apologize studiot. I'm just a little scatter-brained. You kind of have to bear with me on these matters. I'm sure you are used to conversing with some very sharp and knowledgeable people here on this forum, so it might be the case that I test your patience a little more than others when it comes to maths. I usually have to flounder for a while before I start to understand.

As I promised above, I have written down all of the problems on my test review, and here are all the problems arranged in their respective categories:

Differentiate:

y = 6e^x + 4/3√x (the last part is cube root of x)

g(u) = √2u + √3u

g(x) = 5e^x √x

F(y) = (1/y^2 – 3/y^4) (y + 5y^3)

y = x^3/1 - x^2

f(x) = ax + b/cx + d

y = b cos t + t^2 sin t (with respect to t)

y = 6 – sec x/tan x

f(x) = log10 (x^7 + 8)

f(x) + sin (x) ln (8x)

G(y) = ln (4y + 1)^3/√(y^2 ÷ 1)

Find the Derivative:

F(x) = (x^4 + 9x^2 -7)^8

y = cos (a^9 + x^9)

y = 9^(5 – x^2)

g(x) = cosh (ln(x))

y = sin^-1 (4x + 1)

Implicit Differentiation dy/dx:

4x^3 + x^2y + xy^3 = 2

y cos x = 3x^2 + 2y^2

Implicit Differentiation to find equation of tangent line to the curve:

5x^2 + xy + 5y^2 = 11, (1,1) (ellipse)

Find equation of tangent line to the curve:

y = 10x sin x, (π/2,5π)

y = x^4 + 5x^2 – x, (1.5)

Logarithmic Differentiation:

y = √(x – 5/x^6 +4)

y – (cos 8x)^x

Linearization L(x):

f(x) = x^4 + 6x^2, a = -1

f(x) = x^(2/3), a = 8

Find Numerical Value:

Cosh 3

Cosh (ln3)

Word Problems:

Each side of a square is increasing at a rate of 8 cm/s. At what rate is the area of the square increasing when the area of the square is 49 cm^2?

The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing when the diameter is 60mm?

A plane flying horizontally at an altitude of 1 mi and a speed of 410 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station. (round your answer to the nearest whole number.)

I understand you guys cannot work all of these problems or address them all. I was hoping perhaps we could tackle a few (or several of them), at least identifying the concepts that need to be studied. I also posted them all so you could get a feel for the trends in concepts that are being used here and can better direct on what I need to focus my studies on. I appreciate all the help I can get, and all the help so far!

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Thanks for reminding me ajb, I was going to say that as well

+1 Klaynos

Tampitump, your approach now should very much depend upon the sort of questions you are likely to be asked in your calculus exam.

Do they want you to show understanding of the underlying principles?

Do they want you to be able to extract derivatives and integrals according to particular rules and methods?

Do they expect you to apply the knowledge of calculus to anything (eg maxima and minima) ?

Anything else?

Whilst I am suggesting you apply your main focus to your exam requirements, to help us know where you are coming from can you tell us what you already know pre calculus.

Do you for instance know what a function is?

What notation are you using?

There you are then you have answered most of my questions and you now need a crash course in differentiation methods.

Integration is yet to be tackled.

What you have listed in post#14 is basically the algebraic manipulation of derivatives and a few examples of use.

Yes differentiation is basically algebraic manipulation.

There are not many rules, you could probably learn enough to pick up enough marks to scrape by since you obviously understand basic algebra.

(Not a path I'd recommend but it can be done and would be the start of your recovery).

Edited by studiot
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There you are then you have answered most of my questions and you now need a crash course in differentiation methods.

Integration is yet to be tackled.

What you have listed in post#14 is basically the algebraic manipulation of derivatives and a few examples of use.

Yes differentiation is basically algebraic manipulation.

There are not many rules, you could probably learn enough to pick up enough marks to scrape by since you obviously understand basic algebra.

(Not a path I'd recommend but it can be done and would be the start of your recovery).

Okay, thanks. I'm studying right now. I've done some studying today and it would appear that to solve most of these differentiation problems, I need to learn the various rules that are used to solve them (i.e. product rule, chain rule, power rule, quotient rule), and when to use these various methods. I think I've got the power rule down (i.e. the derivative of x^4 would be 4x^3), but there are some problems on here that seem a little trickier. I've also been told that there are shortcuts to differentiating which our professor has said we are about to get into in this course.

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Okay, thanks. I'm studying right now. I've done some studying today and it would appear that to solve most of these differentiation problems, I need to learn the various rules that are used to solve them (i.e. product rule, chain rule, power rule, quotient rule), and when to use these various methods. I think I've got the power rule down (i.e. the derivative of x^4 would be 4x^3), but there are some problems on here that seem a little trickier. I've also been told that there are shortcuts to differentiating which our professor has said we are about to get into in this course.

It is normal to have a table of standard derivatives available in exams, though it is quicker if you know them.

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It is normal to have a table of standard derivatives available in exams, though it is quicker if you know them.

Having the table doesn't guarantee I know how to use them. I'm not sure how to take the derivative of certain expressions. Like √(2u) or √(3u)

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√(x) is the same as x^(0.5). The first step in the ones you cannot do is to convert into that form then use the power rule.

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With that, I add that you should also learn the derivatives of trigonometric functions.

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Having the table doesn't guarantee I know how to use them. I'm not sure how to take the derivative of certain expressions. Like √(2u) or √(3u)

The table only contains the very simple, I'm sure you can cope with

$y = a{x^n}$
$\frac{{dy}}{{dx}} = na{x^{\left( {n - 1} \right)}}$
u and v are universally used to show an intermediate function of x.
This is also known as function of a function or the chain rule, but both are fancy names for substitution.
Watch
$y = \sqrt {3{x^2}}$
$Let\;u = 3{x^2}$
$y = \sqrt u = {u^{\frac{1}{2}}}$
to differentiate split the derivative into two parts
$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}*\frac{{du}}{{dx}}$
$\frac{{du}}{{dx}} = 6x$
$\frac{{dy}}{{du}} = \frac{d}{{du}}\left( {{u^{\frac{1}{2}}}} \right) = \frac{1}{2}{u^{\left( {\frac{1}{2} - 1} \right)}} = \frac{1}{2}{u^{ - \frac{1}{2}}} = \frac{1}{{2\sqrt u }} = \frac{1}{{2\sqrt {3{x^2}} }}$
$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}*\frac{{du}}{{dx}} = \frac{1}{{2\sqrt {3{x^2}} }}*6x = \frac{{3x}}{{\sqrt {3{x^2}} }} = \frac{3}{{\sqrt 3 }} = \sqrt 3$
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Ok, I think I've got the product, chain, quotient, and power rules down pat (at least their methodologies). I had some help from my school's online tutor service this morning and that was really helpful. Its pretty easy after you practice a few examples. There are some situations I have trouble with though. I will post some examples later. I have some work to get done for another class right now so I will check back in later.

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Careful.

You might get to like some of this stuff.

Edited by studiot
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Welp, I failed my calculus midterm..... MISERABLY! Semester ruined.

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TYPE: SINGLE_QUESTION_WITH_ANSWER

The text passage contains a personal statement about the outcome of a calculus midterm exam, which is a mathematical assessment. The user expresses disappointment in their performance, indicating they failed the exam and feel their semester is now negatively impacted. The question is implicit in the context, as the user is essentially asking for advice or support after this academic setback. The answer is provided through the context of the situation, which is the user's expression of their failure and the impact on their semester. However, the answer is not a direct solution to a mathematical problem, but rather a statement of the user's emotional and academic state.