Quick math question

audiobliss

Hey fellas, me gotst a quick math prublem.

When you have a variable (or anything) squared, for instance, in the denominator, you can move it to the numerator by negating the exponent. My question is, when I move the variable (in this case) to the numerator, am I adding the variable, or multiplying the whole numerator by that variable?

Thanks!!

CrBoy

If you're dividing, when you move up the variable with the neg exponent, you have to multiply the numerator by the denominator... is this clear enough? I hope so

jdhdiggs

Word problems blow...

PolkThug

The answer is 42.

audiobliss

Ok, I Just thought about it for a moment and proved to myself that you have to multiply.

So anyway, I'm trying to take the derivative. I just learned that the derivative = nx^(n-1). So, I'm using that. What I'm trying to find the derivative of is:

x^2 + 4x + 3
____________
sqrt(x)

So, I can say that equals:

x^2 + 4x + 3
___________
x^(1/2)

Then I can move the denominator to the numerator:

(x^2 + 4x + 3)(x^(-1/2))

Distributing:

x^(3/2) + 4x^(1/2) + 3x^(-1/2)

Then finding the derivative (nx^(n-1)):

(3/2)x^(1/2) + 2x^(-1/2) - (3/2)x^(-3/2)

Is that right?

Thanks!!

audiobliss

Nice diagram, crboy! Makes it very clear.

PT - I guess I got it wrong. Not exactly what I came out with.


The underscore is supposed to be the line separating numerator from denominator, in case you couldn't tell. I couldn't think of a better way to do it.

Thanks for the responses!

CrBoy

I'm glad to help...

gatemplin

AB,

That looks right, but in the future just use the quotient rule:

f (x)= a/b

f '(x)=(a'b-ab')/(b^2)

There is less chance of error, especially when the exponents and coefficients get uglier. You wont always be able to simplify first like that.

audiobliss

Hmm. I have no idea what that is. Obviously it's another way to find the derivative, but is it some sort of theorem that I'll come across later in Calc?

So basically you're saying that if your function is a over b (a and be standing for expressions), then the derivative of the function is the derivative of a (the numerator) times b (the denominator) minus a times the derivative of b all divided by the square of b?

That's interesting to know, and if I understand it right, I may play around with it some, but I won't be able to actually use it if we don't learn (or establish) it in class first.

Thanks!

mldennison

ugh, i always hated that, there is a nice simple way to solve a problem in the next chapter but since you "need" to know how to do it the hard way you are not allowed to use that until next week, sorry!

gatemplin

audiobliss wrote:

So basically you're saying that if your function is a over b (a and be standing for expressions), then the derivative of the function is the derivative of a (the numerator) times b (the denominator) minus a times the derivative of b all divided by the square of b?

That's interesting to know, and if I understand it right, I may play around with it some, but I won't be able to actually use it if we don't learn (or establish) it in class first.

Thanks!

Exactly right. Usually u and v symbolize the expressions, but a and b were fresh in my head from some proofs in school today.

Many people dont like to use the quotient rule, and manipulate the equation to avoid it. It is easier to make mistakes that way. Just 10 mins ago I was helping a girl who tried to avoid the quotient rule and got the wrong answer because of it.

unc2701

hmmm... I'd say the opposite- no one remembers the quotient rule after about 2 weeks and it's easier to make a big problem into several small, managable ones.

gatemplin

unc2701 wrote:

hmmm... I'd say the opposite- no one remembers the quotient rule after about 2 weeks and it's easier to make a big problem into several small, managable ones.

Hey...dont be dissin' the quotient rule!