guess i should of wrote my notes clearly in class

find y.

a. y=5x^2-2x+3
b. y=(x^2-2x)(2x^3-x^2)
c y=(x^2-2x)/x-2
d y^2+5x^2=2x+3

any help will do dont even have to give me an answer just need to understand how to do it.

What do you mean find y? Are you trying to take the derivative? Y is already found in terms of x. If you'd like to negate the terms of x, you have to derive Y, but this won't give you a Y, it'll give you a Y'

Derivatives are VERY easy. I'll do the first one for you:

1) Y = 5x^2 - 2x + 3

When you take a derivative, you move the exponent down to the front of the x, and subtract that exponent by 1. I'll show you:

Say you have a number x^n. When taking the derivative, that number becomes nx^(n-1). An example of this: x^6 becomes 6x^5. If the term doesn't have an exponent (6x for instance) you may re-write it as 6x^1. Taking the derivitive of 6x^1 gives you 6x^0. Anything raised to the 0th power however is just 1, so x^0 = 1, so 6x^0 just simply equals 6. Any constants when taking a derivitive are just eliminated (a constant is a variable or number that isn't involved in the derivite. A floating number, like 3, is just nullified. I'll show you in a recap). Quick recap:

Y = 5x^2 + 3x - 2
dY/dx (derivitive, can also be written as Y') = 10x + 3

So back to your problem:
a)Y = 5x^2 - 2x + 3
Y' = 10x - 2

b)y=(x^2-2x)(2x^3-x^2)
There are two ways you can approach this problem. You can foil everything together, or you can use the product rule of calculus derivatives. I'd suggest using the product rule, since you'll need to learn how to use it anyway. The product rule is simple:

Say you have two quantities: (3x^2 + 2x - 3)(4x^3 - 3x - 1)
To do the product rule we first assign each portion a variable:
(3x^2 + 2x - 3) = f
(4x^3 - 3x - 1) = g.

The product rule states Y' = (f' * g) + (f * g').
Here's how that works out for my example (3x^2 + 2x - 3)(4x^3 - 3x - 1):
Y' = (6x + 2)(4x^3 - 3x - 1) + (3x^2 + 2x - 3)(12x^2 - 3)

So for your problem in part b, when you do it you should end up with:
dy/dx (or y' however you wish to write it) = (2x-2)(2x^3-x^2) + (x^2-2x)(6x^2-2x)

c) y=(x^2-2x)/x-2 There is only one way to do this problem: you must use the quotient rule. The quotient rule is also quite simple. Lets use the example Y = 2x/x^2 We first separate the top and the bottom, like so:
g = 2x
h = x^2

Now, we apply the formula for the quotient rule: Y' = (g'*h - h'*g)/h^2
So we have Y' = [(2x * x^2) - (2x * 2x)] / x^4

For c, that gives us an answer of
y' = {[(-4x-2)*(x-2)] - [(x)*(x^2-2x)]} / x^2 + 4

For problem d this is the hardest of all. Since the x's are exponentiated with the y, you must first take a log(base 2+5x^2) of both sides. This gives us
y = log(base: 2+5x^2)*2x+3

Here's a guide on how to derive a logarithmic system: http://answers.yahoo.com/question/in...6174444AABgttU

Hope it helps.

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Your C is off unless you simplified and canceled some stuff out.

On the top its 2x - 2 where you have -4x - 2. Also deriv of x-2 is 1 not x. Lastly (x-2)^2 is not x^2 + 4.

For D, you must be smoking something cause I do not see a reason for any log stuff. There is no y^x stuff there.


Please proof read your stuff and check for mistakes or else you'll just be leading these people down the wrong road.

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y^2+5x^2=2x+3

Now do you see it?

Also, my math is a bit off, sorry, just getting over norovirus. x-1 is just 1 when derived (since the -1 goes away and the x just becomes a 1).

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O wow, I assumed its not part of the exponent because there's no parenthesis. If it is like that, then the problem seems out of context since it suddenly got so much more difficult than the others.

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thanks for the help i think i got the understanding of it now. got a different answer for b i remember the teacher explaining product rule but that was towards the end of the day and my brain was dead by then and c an im lost on totally lost on d. got class today so ill just ask the teacher to explain all that stuff