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[klingon13524]

One word: interest rates.

That's two words Mr. hotshot mathematician.

Congratulations, you got the joke.

[sarcasm] And you got my joke! [/sarcasm]

[ZetaX]

Damn exponents and square roots... Why would I ever need to know this?

Damn words and letters... Why would I ever need to know this¿

[ToonyMan]

Can somebody go through all the steps to solve this problem?  I assume there's more than one way so you can do whatever.  Thanks in advance.

[ed boy]

Nevermind, I was wrong

[ToonyMan]

The back of my text book says the answer is 1/8.  I can't find a way to get there though. 

[Christes]

Multiply the top and bottom by sqrt(16+h)+4.

On the top, you should get just h, which you can cancel with stuff on the bottom.  You should then be able to plug 0 in for h.

[alway]

I think it requires finding the derivative of √(16+h).
√(16+h) = (16+h)^(1/2); assuming I remember my calc correctly, the derivative of that is (1/2)*(16+h)^(-1/2)
as h approaches 0, (1/2)*(16)^(-1/2) = 0.125
That value is the ratio relating the two h values; the √16 and -4 cancel out all but the change resulting from the h, leaving you with 1/8, as the top half approaches zero at a rate of 1/8 that of the bottom.


To work through it with Christes' method:
(√(16+h)+4)(√(16+h)-4) / h*(√(16+h)+4) =
(16+h-16)/h(√(16+h)+4) =
h/h(√(16+h)+4) =
1/(√(16+h) + 4) =
as h approaches 0, 1/(√16+4) =
1/(4+4) = 1/8

[Soadreqm]

Some may consider Wolfram Alpha to be cheating. Click on "show steps".

[Darvi]

That's... the derivative of the sqrt function on x=16 I think.

In other words, (sqrt(x))' where X=16.

I think. You wouldn't believe what you forget in two months.

Spoiler (click to show/hide)

(sqrt(x))'=(2sqrt(x))^-1

You do the math.

In other words, what alway said but simplified.

[Christes]

I guess I just assumed that Toonyman didn't have calculus on hand, since L'Hopital's rule is always the go-to theorem after you learn it

[Darvi]

Well technically when you use hopital then that's exactly what you get. Because the lower part becomes one then you only get the derivative of the top part.

[ToonyMan]

I've seen it in class I just wasn't sure how to multiple it.  Thank you.

[eerr]

I would just substitute b^2= 16 +h

what was the method the teacher used?

[Urist Imiknorris]

FFFFFFFFFFFFUUUUUUUUUUUUUUUUUUUUUUU-

Find a general solution of the following differential equation:
x(dy/dx) = y + √(x² + y²)

What I've got so far:

v = x² + y² <--substitution
y = √(-x² + v)
dy/dx = 1/2(√(-x² + v))(-2x+ (dv/dx)) + 1/2(√(-x² + v))(-2x+1)(dx/dv)
dv/dx = -2x² + 2v +2√(-x² + v)√(v) + 2x + (2x+1)(dx/dv)

a) Which question should I be asking: Where did I go wrong? or What is dx/dv?
b) What's the answer to the question I should be asking?

Fucking differential equations and their eldritch ways.

[Jim Groovester]

I don't follow. What are you trying to do?

Could you post a few of the steps of how you got to those points?