The Energon Pub

[robofreak]

Hey everyone! I'm trying to brush up on my calculus before the semester starts and I've run into a snag on derivatives.

The problem:

f(x)= (3x^2+4x-8)(2x^3-5)

The needed result has to be the derivative. I know for this problem, I need to use the product rule which is: f'(u*v)= u'*v+u*v'

u = 3x^2+4x-8
v = 2x^3-5

Thus:

u' = 6x+4
v' = 6x^2

Knowing this I end up with this when I apply the product rule:

f'(u*v) = (6x+4)(2x^3-5)+(3x^2+4x-8)(6x^2)

f'(u*v) = 12x^4-30x+8x^3-20+18x^4+24x^3-48x^2

Now this is where I hit the snag. I can't remember how to simplify it. My first impulse is this:

f'(u*v)= 30x^4-30x+32x^3-48x2-20

That does not look right to me. Can someone tell me where I went wrong and how to fix it?

[Predaprince]

That is correct, but just rewrite it so it is in the correct order.

30x^4 + 32x^3 - 48x^2 - 30x - 20

That may be why it didn't look right to you.

[robofreak]

That is correct, but just rewrite it so it is in the correct order.

30x^4 + 32x^3 - 48x^2 - 30x - 20

That may be why it didn't look right to you.

Thanks. I thought I was doing it wrong when I was adding the exponents. Looks like I did it right.

I will have more questions ove rht enext couple days though. This course will be the end of me.

[Predaprince]

That is correct, but just rewrite it so it is in the correct order.

30x^4 + 32x^3 - 48x^2 - 30x - 20

That may be why it didn't look right to you.

Thanks. I thought I was doing it wrong when I was adding the exponents. Looks like I did it right.

I will have more questions ove rht enext couple days though. This course will be the end of me.

I teach Honors Algebra and Honors Geometry to eighth graders. It has been six years since I took Differential Equations aka the combined class of what use to be known as the individual classes of Calculus 4, 5, and 6. I am ready to be challenged.

[robofreak]

Okay, we got a new one here. Please ignore the underscore marks. It was the only way I could make the fractions look right.

f(x)=6x^2-5x+1
____3x^2-5

For this I need the quotient rule:

f'(x)= u'*v-u*v'
_____v^2

u=6x^2-5x+1
u'=12x-5

v=3x^2-5
v'=6x

Now with u and v with there respective derivatives declared, I can set up the equation:

f'(x)= (12x-5)(3x^2-5)-(6x^2-5x+1)(6x)
________(3x^2-5)^2

Now I get:

f'(x)= 36x^3-60x-15x^2+25-36x^3-30x^2+6x
_____9x^4-15x^2-15x^2+25

Now here's where I hit the snag. The simplification. I want to do this:

f'(x)= -54x-45x^2+25
____9x^4-30x^2+25

Where did I mess up and what am I missing. It does not look right to me. It's been a while since I've done math so that's also a factor in why it may not look right.

[Predaprince]

I see a very common mistake that my Algebra 1 students constantly make. When you multiplied (12x-5)(3x^2-5)-(6x^2-5x+1)(6x), you forgot that the minus sign needs to be "distributed" to all parts of the second product.

You should now have


36x^3-60x-15x^2+25-36x^3+30x^2-6x
9x^4-15x^2-15x^2+25

15x^2 - 66x + 25
9x^4 - 30x^2 + 25

[Tigertrack]

I see a very common mistake that my Algebra 1 students constantly make. When you multiplied (12x-5)(3x^2-5)-(6x^2-5x+1)(6x), you forgot that the minus sign needs to be "distributed" to all parts of the second product.

You should now have


36x^3-60x-15x^2+25-36x^3+30x^2-6x
9x^4-15x^2-15x^2+25

15x^2 - 66x + 25
9x^4 - 30x^2 + 25

Distribution... indeed a very common error.

And I haven't done this level of math in about 15 years!

[robofreak]

Distribution... indeed a very common error.

And I haven't done this level of math in about 15 years!

Ha! I knew I was nver going to use calclulus again.

Anyways, my course just started so now the string of questions will soon begin.

[Galvatron X]

It has been six years since I took Differential Equations aka the combined class of what use to be known as the individual classes of Calculus 4, 5, and 6...

Yeah, it's been a while for me too! I graduated from The Univerity of Maine (Mechanical Engineering) in 2003.

Anyways, my course just started so now the string of questions will soon begin.

Bring it on!

I've helped out a few people over the past few years with their math. Just two weeks ago, actually. This will be good for me too - I always try to keep that stuff at least somewhat fresh in my mind!

[robofreak]

y= 3(x+10)(x^2-100)

Okay, finding the derivative of this has me lost and I know I'm missing something simple.

The problem I've always had with math is that I have to see how to do the problem first. The notes didn't go over a set up like this so I'm a little lost. Actually, it's that stupid 3 that is throwing me off. I don't know why, but it is.

I'm probably going to feel really silly after seeing how the problem is to be answered too.

[Predaprince]

You are still going to use the product rule. For three functions, the derivative of U*V*W is U'*V*W + U*V'*W + U*V*W'.

y' = (3)'*(x+10)*(x^2-100) + 3*(x+10)'*(x^2-100) + 3*(x+10)*(x^2-100)'

[robofreak]

Sorry I didn't thank you earlier Predaprince. life is hectic with school.

Anyways, since solving this, it keeps nagging at the back of my head and I finally figured out why. My mind wanted to over complicate this when I all I had to do was this:

y=3(x+10)(x^2-100)

y=3(x^3-100x+10x^2-1000)

y=3x^3-300x+30x^2-3000

Now I can just take the derivatives straight across.

y'=9x^2+60x-300

Granted, this is a more algebraic way of doing it compared to the product rule, but it seems a little less messy when you just do it like this.

I'm steadily learning more and more to use algebra as much as possible before I make the switch to doing calculus. Keeps things cleaner and a lower margin of error it seems.

Is my logic flawed in this thinking?

[Predaprince]

You are also correct. By multiplying out the original function and then getting the derivative, it is much simpler for that problem and what you have is the correct answer.

[robofreak]

Does anyone have any tips for doing Gaussian Elimination?

I know that there is a method for doing it in my calculator, but I can't figure it out. I'm using a TI-84.

can anyone give me a step by step? Here's an example:

7x + 6y = -1
6x + 7y = 1

I believe the proper term I'm looking for is row reduction. I bascially have to find x and y for both of them.

[Predaprince]

7x + 6y = -1
6x + 7y = 1

Multiply the first one by -6 and the second by 7. Then combine to eliminate the x terms and solve for y. Once you have what y equals, substitute it into one of the original equations to solve for x.

[Wheelimus Prime]

@Predaprince:

so, i didn't come back to seibs for math help,

but i would like you to know you helped me out with an Alg 2 problem of mine.

thanks for the thread, Robofreak

[prowl123]

Jesus! And I thought Algebra 1 was hard!

[robofreak]

Jesus! And I thought Algebra 1 was hard!

Just wait until my questions about multivirate calculus start up.

[prowl123]

Jesus! And I thought Algebra 1 was hard!

Just wait until my questions about multivirate calculus start up.

Huh?

[robofreak]

I need someone that is really good at calculus to PM or email me at robofreak@seibertron.com

I could use some assisstance by getting some walkthroughs in some work.

I normally wouldn't ask this, but the tutor I've been seeing is out of town so I could really use the aid of someone that can help me out for a bit through a skype call or chat box.

I look forward to hearing from you.

I need help within the next couple of days. Hopefully we can work something out.

[prowl123]

Wait a minute... are the ^ supposed to represent exponents?

If so, then:

y=3(x+10)(x^2-100)

y=3x+3(10)(x^2-100)

y=3x+30(x^2+(-100))

y+3x+x^2-70

I have no clue how the rest goes. That's my Algebra brain going crazy.

[Predaprince]

y=(3x+30)(x^2-100)
y= 3x(x^2-100) + 30(x^2-100)
y= 3x^3 - 300x + 30x^2 - 3000
y= 3x^3 + 30x^2 - 300x - 3000
the derivative of this is y' = 3*3x^(3-1) + 2*30x^(2-1) - 1*300x^(1-1) = 9x^2 + 60x - 300