Live from Carbondale, it's Tuesday night! Welcome to your first scribe post. Our first scribe post is about the product rule, and since I can't think of any cool pictures that would represent the product rule, I'm just going to jump right in.

We started exploration 4-2 in class on Monday with this objective: Given a function that is a product of two other functions, find in one step, an equation for the derivative function. In math terms: if f(x)=g(x)*h(x), what will f'(x) be? First, we figured out if differentiation distributes over addition through this test:

Using a function that was a sum of two functions,:

y=(x

^{3})+(5x+1)we first differentiated by adding the derivative of each separate function:

d/dx(x

^{3})+d/dx(5x+1)this equalled:

=

**(3x**^{2}**)+(5)**Then we differentiated the function as a whole:

d/dx(x

^{3}+5x+1)this equalled:

=

**(3x**^{2}**+5)**

As you can see, the derivative of a sum of two funtions is equal to the sum of the derivatives of the two functions. This is how we discovered that YES, differentiation does distribute over addition. The we wondered, does it distribute over multiplication? We used similar steps to answer this question:

we started with the same composition of equations, but this time they were multiplied:

y=(x

^{3})*(5x+1)we differentiated each equation seperately:

d/dx(x

^{3})*d/dx(5x+1)this equalled

(3x

^{2})*(5)which simplifies to:

**15x**^{2}then we wanted to differentiate as a whole equation, so first, we simplified by distributing the x

^{3 }to the 5x+1 which equalled:5x

^{4}+x^{3}then we differentiated the equation as a whole:

d/dx(5x

^{4}+x^{3})This equals:

**20x**^{3}**+3x**^{2}

As you can see from this example, the derivative of a product of two functions is __not__ equal to the the product of the derivatives of the two functions. This is how we discovered NO differentiation does NOT distribute over multiplications. So then we knew that we needed to find a way to find the derivative of a product of two functions. We then used the definition of derivative to derive the formula for the derivative of a product of two functions. Below is a slide from Bru's presentation that has the answers to the proof on the second page of your exploration. Below is an explanation of each step.

I'm really sorry if you can't read this, the same thing can be found of the slide show if the product rule that Bru posted.

__1.-2.:__ Since y=uv, Δy=ΔuΔv and

(y+Δy)= (u+Δu)(v+Δv)

__2.-3.:__ FOIL

__3.-4.:__ The positive *uv* at the beginning and the negative *uv *at the end add to 0.

__3.-4.:__ explained

__4.-5.:__ You want the Δu or Δv to be in the fraction and the u or v to next to it. Bru did most of this step for you.

__5.-6.:__ Explained, take the limit of all three seperately.

__6.-7.:__ Δ symbol changes to *d* to symbolize that the limit is a derivative. The Δ*u* in the third becomes zero based on the graph at the top of the page on the far left. We are taking the limit as Δ*x* approaches 0 and as you can see by this graph, as Δ*x* approaches 0, Δ*u* also approaches 0.

So, the product rule is:

If f(x)=g(x)h(x)

Then f'(x)=g'(x)h(x)+g(x)h'(x)

We ended class with a couple of practice problems and learned that the algebra part (simplifying) is actually the hardest part. Here is one example we did in class. My commentary and additions are in red.

Check Your Understanding:

If f(x)=[(3x-8)

^{7}][(4x+9)^{5}], find f'(x)I put in the red brackets to help me figure out what the two different functions are.

f'(x)= (7(3x-8)^{6}•3)((4x+9)^{5})+((3x-8)^{7})•(5(4x+9)^{4}•4)

Take a look at the first term. In this term, you differentiated the function (which happened to be a composite functions and required the chain rule), and multiplied it by the second function. Then in the second term, you took the first function and multiplied it by the derivative of the second function (which was also composite). As long as you can keep all this straight in your head, this step is relatively easy. But now you have to simplify.

f'(x)= (3x-8)^{6}(4x-9)^{4}[21(4x+9)+20(3x-8)]

In this step you factored out (3x-8)^{6 }and (4x-9)^{4}. Then, you were only left with what is inside the brackets.

f'(x)= (3x-8)^{6}(4x-9)^{4}[84x+189+60x-160]

Distribute within the brackets.

f'(x)= (3x-8)^{6}(4x-9)^{4}[144x+29]

Combine like terms within the brackets, and now the equation is fully simplified.

My answers for numbers 3, 4, and 5. Let me know if you agree or disagree.

e

^{sinx}cos^{2}x-e^{sinx}sinxx

^{-6.3}(1/x+ln4x*-6.3x^{-1.3})(21x

^{6}-60x4)cos10x-10(3x^{7}-12x^{5})sin10xI think this one can be simplified more. I tried to factor out 3x

^{4}but it didn't work, or maybe I did it wrong.

Enjoy! Comment if you wish.

Next Scribe is Babar.

Hello Secret,

ReplyDeleteMy answers for #'s 3,4, and 5 were very different, I am not sure of my algebra so I was wondering if next time you could put in your post the details and steps you went through to get your answers, otherwise great post, this is really helpful to go over before class!

Secret, great first scibe post. You have set the bar of excellence for future scribe posts.Your summary of the class was complete and easy to follow. I liked your use of color to highlight key points in your post. It was especially effective in pointing out the two functions used in Problem 2. Your explanations for each step in the derivation of the Product Rule will help your classmates understanding. I enjoyed the conversational tone of your post. I very much appreciated that you ended the post with a question to your classmates to check your answers to the homework problems Well done!

ReplyDeleteHey Secret! From the beginning to the end, I was very amazed by everything you did. Language, colors, explanations,etc. everything is thorough and easy to understand. The thing I liked the most is your comments/explanations in the "check your understanding" part. You took your time to describe all the steps, even the concepts we learned last semester and the years before (chain rule and factoring.)

ReplyDeleteI really likes the fact that you explained everything in your own language.

Great job. Keep this up. :)

This scribe post was thorough yet very easy to follow. This was a great followup to the class itself. I found that the color coding helped me to realize when I was reading something important, which is very meaningful to me because often times in class I have trouble prioritizing what is the most important. I also found it very helpful how you posted the explanations to each step of derivation in your own words, which made the seemingly daunting process user friendly. I got the same answers as you did for the last questions. Bravo

ReplyDeleteHey secret (and class),

ReplyDeleteI wanted to share a little device I use for "keeping it all straight in my head" when I'm doing the product rule. I think of the original functions in the product (your g and h) as functions 1 and 2. Then the product rule is 1'2+12' (I know its a little nonstandard to name functions with numbers, and I don't actually write this down anywhere), but then I can sort of sing in my head "one prime two plus one two prime" which is a lot shorter than "g of x prime times h of x plus g of x time h of x prime" or some such. Make sense? Let me know if this is complete nonsense to you.

Also wanted to point out what looks like an excellent use of the equation editor.

BEAUTIFUL. Great first comment, way to take the lead. i love the use of color it made things really easy to see and defferentiate different parts of the problem. What helped me a lot was seeing some help on the homework, i always second guess myself and and change my answer to the wrong answer, so it was nice to see someone else's work to reinforce my confidence in answering questions. Thank You.

ReplyDeleteI tried really hard to find some mistakes in this post. However, I couldn't find any. Awesome work!!!!!!!

ReplyDeleteThis post includes everything we talked about in the class. What is really neat of this post is that a lot of details were provided. That's really nice. And I actually understand the product rule better after reading this post.

Thanks a lot!!!

One detail that I want to point out here is the way of typing math equation. It is understandable. However, it is not clear.

For example,

your answer for question number 4.

Just look at this part of the equation "ln4x*-6.3x". I think you see what's wrong.

At last, I have a quick question about the product rule.

Right now, we only know how to deal with the product of two functions.

How to apply the product rule in a product of four functions situation? And you cannot combine any two functions to one.

It will be easy to understand if i just give you an example.

f(x)=(ln4x)(e^sinx)(cos(2x+1))(x+1)^2

f'(x)=?

Everybody give a try here.....

Thank you all for the wonderful comments, I'm glad you enjoyed my post. kwad, thanks for the extra tip on the product rule, that definately is an easier way to think of it and I'm sure that it will help everyone. YDplusSB, I'm not sure I know what you mean about the way I typed my equations. And we didn't learn about more that one eqution during the class I scribed for which is why it was not in my post. We did go over it briefly in a later class and I think the first course of action would be to try to combine the equations into two and then use the product rule from there.

ReplyDeleteWay to start off the Scribe Posts with an excellent start. I like how you began the post with a semi- informal, conversational tone as if starting a class. The usage of different colors to help accentuate the concepts you saw as important as well as the answers to the homework problems helped me fully understand the Product Rule. You clearly went above and beyond with your explanations. Great job and thank you for explaining each step of the proof. That really helped me as well!

ReplyDeleteVery good job secret. It aint easy to be the first one up but you nailed it. I liked how you put important phrases and words in color. Nice idea to put commentary in red. It made the problems a lot easier to follow. Your proof was also really understandable. And nice SNL reference

ReplyDelete