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Discussion in 'Community Discussion' started by TSE, Oct 27, 2010.
God I feel stupid.
I am solving derivatives, so how do I simplify 2(2+h)^3?
I think you mean expand. That is already simplified.
I recommend you check the answer since I did that in my head and on my iPod touch. Also do not freak out that I did it in my head because I was applying some short cuts that I know. Short cuts you should not be using in your work. Do not expect your teacher to teach it to you because understanding the short cut requires much higher level of math than you more than likely will ever go. I say this because very few people do enough math to minor in it.
First, you always bring the exponent down and then subtract it by one and use the chain rule.
So it would be d/dh(2(2+h)^3) = 2*d/dh(3*(2+h)^2*d/dh(2+h))
What I did was bring the 3 down and multiply it by the rest and subtract the exponent by one. So that's why it's squared. Next, you use the chain rule and that says you need to take the derivative of the inside so d/dh(2+h).
So the answer is 6*((2+h)^2)
Basically you know how you take the derivative of 6x^3? You bring the exponent down and subtract it by one. So it would be 3*6x^2. This is basically the same thing but since you have a quantity, you have to use the chain rule.
That's cheating you have to use 1st Principles
Awwww but that one takes too long to do
And he's probably going to learn the short way of doing it anyways eventually
No, he meant the derivative. Expanded, it is 16 + 24h + 12h^2 + 2h^3, which is not even close to what you wrote.
The derivative of that is 24 + 24h + 6h^2, which yields 6(h+2)^2, or the same answer as a previous post via the chain rule.
Transform it into the s-plane using LaPlace and then find the Transfer function and the Closed-Loop Transfer Function assuming H(s) is 1. Finally, find the Sensitivity with respect to T, the damping ration and peak time.
Finally, after all said and done, find the output y(t) if r(t) is 10t. Once you get all that, your remaining answer is in simplest form possible.
Jokes over, just expand that function; bonus points for anyone that knows what subject matter I just did up there.
No, you dont just expand the function, unless you go to Texas Tech.
And you used LaPlace transforms in your joke.
There is more to that than just LaPlace, there is an entire course in relation to that. No it was not Differential Equations or another Math Department based course.
Edit - Also, just use the regular differentiation when dividing. d/dh
Yeah that is what I get for trying to do that at 4:30 in the morning from my iPod after I just woke up along with trying to use a fancy mathmatical proof I know. All were just all sorts of things asking for trouble.
Now the derivative would of been easy provided I was not required to do it with limit definition. But that is what the easy method is for
Posted from my blackberry
+10 Internets for you.
My class name is Classic Control Systems and Design; ENEE 3533.
this one has been made needlessly complicated by some of you--it is a matter of following the order of operations "PEMDAS" first then doing the derivative.
2(2+h)^3 since nothing inside the parentheses can be changed we skip to the exponent placing the exponent inside the parentheses with the two numbers/terms getting:
2(2^3 + h^3) now 2^3 next we see what we can multiply inside the parentheses it can be figured out as 2*2*2 and h^3 remains what it is getting you:
2(8 + h^3) now you Multiply each inside the parentheses with the 2 outside getting you:
2*8 + 2h^3
Then you can figure out the limits to the function yourself I cannot remember the exact calculus at the moment concerning derivatives it's been a while since I had to use any of that.
While you're right about expanding and then taking the derivative, you made a huge mistake. (2+h)^3 is not the same as 2^3+h^3
OH! I see how I screwed that up sorry about that LOL
I know I had to distribute the exponent but I screwed up and forgot about how to do the(a+b)(a+b)(a+b)thing LOL
See I did say I was out of practice--my Instructors in college always said if you forget one rule it can throw off the whole equation even if the rest of what you have written out is entirely technically correct following the steps
Thanks for the help guys!
A lot of knowledgeable and intelligent members here.
I think I did rather well on the test on friday, not complete sure though.