The Energon Pub

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?

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.

Predaprince wrote: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.

robofreak wrote:

Predaprince wrote: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.

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.

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

Predaprince wrote: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!

tigertracks 24 wrote: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.

Predaprince wrote: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.

robofreak wrote:
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!

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.

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)'

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?

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.

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.

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.

@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

Jesus! And I thought Algebra 1 was hard!

prowl123 wrote:Jesus! And I thought Algebra 1 was hard!

Just wait until my questions about multivirate calculus start up.

robofreak wrote:

prowl123 wrote:Jesus! And I thought Algebra 1 was hard!

Just wait until my questions about multivirate calculus start up.

Huh?

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.

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.

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