Slope: The Derivative of a Function at a Point

[morrowcosom]

Homework Statement

We are calculating the slope of the function f(x) = 1/x at x = 1.
--------------------------------------------------------------------------------
So far, so good.
For the function f(x) = 1/x, we now know:

f(1) = 1
f(1+h) = 1/(1 + h)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
f(1+h) - f(1)=
--------------------------------------------------------------------------------

Homework Equations

The Attempt at a Solution

{1/(1+h)-1}/h
{1/(h)}/h
{(1/h)}/h
{(1/h^2)}/h
{h(1/h)}/h
=1/h

I am doing independent study using a computer program, and the program says I am wrong. I looked at an example of these types of problems involving inverses on the website and could not for the life of me understand what I am supposed to do. What in the heck do I do to this problem to solve it?

 

[Mark44]

Homework Statement

We are calculating the slope of the function f(x) = 1/x at x = 1.
--------------------------------------------------------------------------------
So far, so good.
For the function f(x) = 1/x, we now know:

f(1) = 1
f(1+h) = 1/(1 + h)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
f(1+h) - f(1)=
--------------------------------------------------------------------------------

Homework Equations

The Attempt at a Solution

{1/(1+h)-1}/h
{1/(h)}/h

How did [1/(1 + h) - 1]/h turn into (1/h)/h?

{(1/h)}/h
{(1/h^2)}/h

And how did (1/h)/h turn into (1/h^2)/h?

{h(1/h)}/h

(1/h^2)/h isn't equal to h(1/h)/h.

=1/h
I am doing independent study using a computer program, and the program says I am wrong. I looked at an example of these types of problems involving inverses on the website and could not for the life of me understand what I am supposed to do. What in the heck do I do to this problem to solve it?

I found three errors in your work, which suggests that you should review your basic algebra, particularly arithmetic operations on fractions. If you don't have a good understanding of the fundamental algebra operations, you are going to continue having problems with calculus.

 

[morrowcosom]

I have a grasp on arithmetic operations on functions. I have answered 15 questions on the lesson by my self and needed help on about five of them. I just had no idea how to find this answer, so I plugged in everything thing I could possibly think of in order to answer this problem. I just had no idea how to get h by itself on one side, so, I just did what I had to do it, even it was irrational.

If I tried to work {1/(1+h)-1}/h rationally, I would have no idea how to get beyond this point by performing arithmetic in order to eliminate h as a denominator.

The next step I could think of would be to find the inverse of (1/(1+h)) which is (1+1/h), which leaves me with {(1+1/h)-1)/h, which would just lead back to the original problem. I have no idea how to get rid of the 1, and I have no idea how to get h by itself.

I have never come across arithmetic operations on functions where I had to do something like this.

 

[Dick]

It's just algebraic simplification. You've got (1/(1+h)-1) in the numerator. Simplify that first, find a common denominator and combine the two terms.

 

[morrowcosom]

How do you simplify that? I have never ran across anything like that that I had to simplify.

 

[Buri]

Well its like simplifying

(a/b) - 1

You would get 1 as a fraction over b (i.e. get a common denominator of b). So 1 = b/b, so we'd have:

(a/b) - (b/b)

They have the same denominator so we can subtract the numerators now,

= (a - b)/b

You'd basically do the same above..

 

[morrowcosom]

So:

(1/(1+h)) - ((1+h)/(1+h)) = 1-(1+h),


but where do I go from here,

pertaining to my original problem of:
f(1) = 1
f(1+h) = 1/(1 + h)

Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
{f(1+h) - f(1))/h=
I am still left with (1- (1+h))/h

 

[Dick]

For one thing, (1/(1+h)) - ((1+h)/(1+h)) = (1-(1+h))/(1-h). NOT (1-(1+h)). You can't just drop the denominator. For another thing 1-(1+h) is equal to -h, isn't it?

 

[morrowcosom]

Now that I have -h/(1+h)/h, what do I do from here to get rid of the h denominator?
The inside h is an inverse and the outside is not. The only thing I can think of would be to multiply both sides by h to end up with -(h^2)/(1+h).

 

[Mark44]

Now that I have -h/(1+h)/h, what do I do from here to get rid of the h denominator?
The inside h is an inverse and the outside is not. The only thing I can think of would be to multiply both sides by h to end up with -(h^2)/(1+h).

What you started with was
$$\frac{\frac{1}{1 + h} - 1}{h}$$

You can get rid of that bottom h by realizing that dividing by h is the same as multiplying by 1/h. You can rewrite the above as
(1/h)(1/(1 + h) - 1)
= (1/h)(1/(1 + h) - (1 + h)/(1 + h))
= (1/h)((1 - 1 - h)/(1 + h))
= (1/h)(-h/(1 + h)) This is the same as what you have above, but considerably simpler to comprehend.

Using the properties of fraction multiplication, you can simplify this last expression so that you can take the limit as h --> 0. That will give you f'(1), the derivative of the function at x = 1.

 

[morrowcosom]

Thank everyone for their help. When it comes to math, I have no common sense, but want to get good at it.

On:
(1/h)(1/(1 + h) - 1)
= (1/h)(1/(1 + h) - (1 + h)/(1 + h)) How did you get from here
= (1/h)((1 - 1 - h)/(1 + h)) to here
= (1/h)(-h/(1 + h))

I am working a problem now where:
(1/h)(1/(2+h))-1/2
and am working it as

(1/h) (1/((1/2(2 + h)) - ((1/2(2 + h))/((1/2(2 + h))
and getting (1/h) (1-(1/2(2+h)))/ (1/2(2+h))
but the the computer program says the answer is wrong, what gives?

 

[Mark44]

Thank everyone for their help. When it comes to math, I have no common sense, but want to get good at it.

On:
(1/h)(1/(1 + h) - 1)
= (1/h)(1/(1 + h) - (1 + h)/(1 + h)) How did you get from here
= (1/h)((1 - 1 - h)/(1 + h)) to here

In the larger expression there are two fractions with the same denominator. The basic idea is that A/B - C/B = (A - C)/B. This is one of the things I referred to before that you need to brush up on.

I'm going to focus on just part of the expression. I wouldn't normally break out part of an expression, but doing so might help make it clearer what I did.
(1/(1 + h) - (1 + h)/(1 + h)) = [1 - (1 + h)]/(1 + h) = (1 - 1 - h)/(1 + h). This part simplifies to -h/(1 + h).

= (1/h)(-h/(1 + h))

I am working a problem now where:
(1/h)(1/(2+h))-1/2

Judging by what you have below, you don't have enough parentheses. It should be
(1/h)(1/(2+h) - 1/2)

and am working it as

(1/h) (1/((1/2(2 + h)) - ((1/2(2 + h))/((1/2(2 + h))

This doesn't look right, and for sure it's much more complicated than it needs to be. You need to get a common denominator for the expressions 1/(2 + h) and 1/2. That denominator is 2(2 + h). It's probably most helpful if I use LaTeX to represent this work.

$$(\frac{1}{h})(\frac{1}{2 + h} - \frac{1}{2})$$
$$= (\frac{1}{h})(\frac{2*1}{2(2 + h)} - \frac{(2 + h)*1}{2(2 + h)})$$

Now I have a common denominator, which I got be multiplying each fraction by 1 in some form (by 2/2 for the first fraction, and by (2 + h)/(2 + h) for the second).

Continuing,
$$= (\frac{1}{h})(\frac{2 - (2 + h)}{2(2 + h)} )$$
$$= (\frac{1}{h})(\frac{- h)}{2(2 + h)} )$$

All that remains to do is to simplify this and take the limit as h --> 0.

and getting (1/h) (1-(1/2(2+h)))/ (1/2(2+h))
but the the computer program says the answer is wrong, what gives?

 

[morrowcosom]

All that remains to do is to simplify this and take the limit as h --> 0.

I took the answer you found for me and simplified it as follows:

(1/h) / (-h/(4 +2h))
to end up with
-h / (4h +2h^2)

I then set h to 0 and got an answer for the equation of 0, but the computer program says that it is incorrect, where did I screw up, or is the computer program wrong?

 

[Mark44]

I took the answer you found for me and simplified it as follows:

(1/h) / (-h/(4 +2h))
to end up with
-h / (4h +2h^2)

Just because you can multiply two things, that doesn't mean you should do that.
In the fraction on the left there is a factor of h in the denominator. In the fraction on the right, there is a factor of h (times -1) in the numerator.

Before taking the limit, remove common factors; i.e., factors that are present in both the numerator and denominator.

I then set h to 0 and got an answer for the equation of 0, but the computer program says that it is incorrect, where did I screw up, or is the computer program wrong?

Because you didn't simplify the expression first, what you really got when you took the limit was 0/0, which is indeterminate.

 

[morrowcosom]

Even when I simplify down to the simplest point I always get (1/h)(n/d), but in some of these problems I am working, I have to actually divide the (n/d) by (1/h) according to the computer program I am using in order to get the correct answer, even if the division sign does not show up in my work.

Example:

Where I have to divide even though it is not in the work:

We are calculating the slope of the function f(x) = x/(x + 1) at x = 0.
--------------------------------------------------------------------------------
For the function f(x) = x/(x + 1), we now know:

f(0) = 0
f(0+h) = h/(h + 1)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
(f(0+h) - f(0))/h

(1/h)((h/h+1)-(0/h+1))
(1/h)(h/(h+1))
My answer: (h/(h+1) because the two h's cancel out and the 1/h becomes 1/1
The computer's answer: 1/h(h/(h+1)) Why am I having to divide when its is not in my work?

 

[Mark44]

Even when I simplify down to the simplest point I always get (1/h)(n/d), but in some of these problems I am working, I have to actually divide the (n/d) by (1/h) according to the computer program I am using in order to get the correct answer, even if the division sign does not show up in my work.

Example:

Where I have to divide even though it is not in the work:

We are calculating the slope of the function f(x) = x/(x + 1) at x = 0.
--------------------------------------------------------------------------------
For the function f(x) = x/(x + 1), we now know:

f(0) = 0
f(0+h) = h/(h + 1)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
(f(0+h) - f(0))/h

(1/h)((h/h+1)-(0/h+1))
(1/h)(h/(h+1))
My answer: (h/(h+1)

This is incorrect. The h in the denominator of the first fraction cancels with the h in the numerator of the second fraction (as you say below), so
(1/h)(h/(h+1))
= 1/(h + 1)
This is the simplified form of the difference quotient [f(0 + h) - f(0)]/h.
The next step is to take the limit as h --> 0 to get the value of the derivative at x = 0.

because the two h's cancel out and the 1/h becomes 1/1
The computer's answer: 1/h(h/(h+1)) Why am I having to divide when its is not in my work?

1/h does not become 1/1, so I don't know where you're getting that. This expression 1/h(h/(h+1)) is the difference quotient, but not in its most simplified form. That would be 1/(h + 1), as already explained above.

 

[morrowcosom]

I have learned a lot so far, but am having trouble with a different type of problem now:

We are calculating the slope of the function f(x) = -2/x at x = 3/2.
--------------------------------------------------------------------------------

For the function f(x) = -2/x, we now know:

f(3/2) = -4/3
f(3/2+h) = -2/(3/2 + h)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
(f(3/2+h) - f(3/2))/h
--------------------------------------------------------------------------------
My work:
(1/h) (-2/(3/2+h)) + (4/3)
(1/h) (-6/ (3)(3/2+h) + (4)(3/2+h)/ (3)(3/2+h)
(1/h) (4h)/ (3(3/2+h))
= 1/ (3h)/(3(3/2+h))
The website I am using says this is wrong and I have no idea why? Could someone please enlighten me?

Also what is a good website dealing with calculus that gives various examples of each concept. The website I am at now (COW: math at temple) just gives me a few examples in the help section that do not really help on these more complex problems. Going from advanced algebra (the high school early college stuff, not the more abstract stuff) to calculus is in comparison like going from addition to advanced algebra.

 

[Mark44]

I have learned a lot so far, but am having trouble with a different type of problem now:

We are calculating the slope of the function f(x) = -2/x at x = 3/2.
--------------------------------------------------------------------------------

For the function f(x) = -2/x, we now know:

f(3/2) = -4/3
f(3/2+h) = -2/(3/2 + h)
Now evaluate the difference quotient, simplifying as much as possible and cancelling h in the denominator:
--------------------------------------------------------------------------------
(f(3/2+h) - f(3/2))/h
--------------------------------------------------------------------------------
My work:
(1/h) (-2/(3/2+h)) + (4/3)
(1/h) (-6/ (3)(3/2+h) + (4)(3/2+h)/ (3)(3/2+h)
(1/h) (4h)/ (3(3/2+h))

Your work is correct up to here (above), which shows some improvement that I'm glad to see!
Since you are working with so many fractions, it would be worthwhile for you to learn how to format them in LaTeX, which would make them much easier to read, and might eliminate some of the mistakes you have been making. Click on the expression I wrote, below, to see my LaTeX script.

Also, since you are working toward and answer that you plug into a Web page, it's not grading you on your work, but when you get in an actual math class, your instructor might be more picky about such things as connecting equal expressions with an = sign.
The last expression above is
$$\frac{1}{h}\left(\frac{4h}{3(3/2 + h)} \right)$$

The next line should start with =, as should the 2nd and 3rd lines above. The line below is incorrect. From the expression above, there is a factor of h in the denominator of the first fraction and a factor of h in the numerator of the second fraction. Cancel them to get the simplified form of the difference quotient.

After that, take the limit as h --> 0 and you'll be done.

= 1/ (3h)/(3(3/2+h))
The website I am using says this is wrong and I have no idea why? Could someone please enlighten me?

Also what is a good website dealing with calculus that gives various examples of each concept. The website I am at now (COW: math at temple) just gives me a few examples in the help section that do not really help on these more complex problems. Going from advanced algebra (the high school early college stuff, not the more abstract stuff) to calculus is in comparison like going from addition to advanced algebra.

 

[morrowcosom]

The reason I went from
= (1/h) (4h)/ (3(3/2+h))
to
= 1/ (3h)/(3(3/2+h))
Is because I thought that I was canceling the denominator of (1/h) and as a result only able to cancel one h out of the other numerator 4h which left me with 3h as that numerator.

I am confused as how to simplify the like h terms in this problem.


Also, where do I actually input my data so that it can be displayed in a latex format?

 

[Mark44]

The reason I went from
= (1/h) (4h)/ (3(3/2+h))
to
= 1/ (3h)/(3(3/2+h))
Is because I thought that I was canceling the denominator of (1/h) and as a result only able to cancel one h out of the other numerator 4h which left me with 3h as that numerator.

You are confused about how cancelling works. As I suggested before, it would be a good idea for you to spend some time with your old algebra book (or get another one if you don't have that book any more) and review the operations with fractions and rational expressions.
$$\frac{1}{h} \cdot \frac{4h}{3(3/2 + h)} = \frac{h}{h} \cdot \frac{4}{3(3/2 + h)}= \frac{4}{3(3/2 + h)}$$

What you apparently did was to subtract h from 4h to get 3h.

I am confused as how to simplify the like h terms in this problem.


Also, where do I actually input my data so that it can be displayed in a latex format?

I do my LaTeX by hand, although I think there are some tools that have a nice user interface. Here are some links to Web pages about using LaTeX.
http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands
http://heather.cs.ucdavis.edu/~matloff/LaTeX/LookHereFirst.html
http://andy-roberts.net/misc/latex/latextutorial9.html

A LaTeX expression or equation begins with a [ tex] tag and ends with a [ /tex] tag. I have an extra space here so that the tags will be visible.

For a fraction, use \frac{}{} -- inside the tex tags. The expression inside the first pair of braces is the numerator, and the expression inside the second pair is the denominator.
Here's an example. To make it render correctly remove the extra space in the tex and /tex tags.
[ tex]\frac{4}{3(3/2 + h)}[ /tex]

Here it is with the corrected tex and /tex tags:
$$\frac{4}{3(3/2 + h)}$$

If you click the expression above, a new window opens that shows the LaTeX code.

You can do lots of things with LaTeX. For each example I show the LaTeX code as it would look (but with an extra space in the tex and /tex tags to keep it from rendering).
exponents: [ tex]2^{x + 3}[ /tex]
$$2^{x + 3}$$

subscripts:
[ tex]a_1 + a_2[ /tex]
$$a_1 + a_2$$

limits:
[ tex]\lim_{h \to 0} \frac{f(a + h) - f(a)}{h} [ /tex]
$$\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$

derivatives:
[ tex]\frac{d}{dx}x^n = nx^{n - 1}[ /tex]
$$\frac{d}{dx}x^n = nx^{n - 1}$$

integrals:
[ tex]\int 3x^2 dx = x^3 + C [ /tex]
$$\int 3x^2 dx = x^3 + C$$

[ tex]\int_0^2 3x^2 dx [ /tex]
$$\int_0^2 3x^2 dx$$

And lots more...