Damn, I suck at math..... help me out.

[notfred]

OK, so I'm taking calc 2..... and I got stuck on stupid algebra on my first homework assignment... and it's review from my last class.... ugh.

g(x) = sqrt(1+2x)

Find g'(x) using the definitoin of derivative

So: g'(x) = lim as h->0 (sqrt(1+2(x+h)) - sqrt(1+2x))/h

How the hell do I factor an h out of the top of that equation? I haven't done any algebra in a long time. And no I can't just use the chain rule.

 

[rgwalt]

Wow, I guess I suck at math too... here is a bump for you.

Ryan

 

[LakerGod]

 

[Hanpan]

Multiply by


(sqrt(1+2(x+h)) +sqrt(1+2x))/(sqrt(1+2(x+h)) +sqrt(1+2x))

this should give you 1+2(x+h)-1-2x over some stuff

this simplifies to 2h/h*(sqrt(1+2(x+h)) +sqrt(1+2x)

2/sqrt(1+2(x+h)) +sqrt(1+2x)


and Voila 1/sqrt(1+2x)

Edit. Left out a parenthesis

 

[Zorro]

wd40 over V8 = g8

 

[Tommy2000GT]

Ti-89

 

[amdskip]

notfred mode <on>: do your own homework for once so you actually learn something </off>

I took calc 2 but that was too long ago. Sorry I can't help you.

 

[brtspears2]

Hmm, I guess I'm equally bad, I cant figure it out using def of derivative, but chain rule, no problem at all.

 

[notfred]

Originally posted by: Hanpan
Multiply by


(sqrt(1+2(x+h)) +sqrt(1+2x))/(sqrt(1+2(x+h)) +sqrt(1+2x))

this should give you 1+2(x+h)-1-2x over some stuff

this simplifies to 2h/h*(sqrt(1+2(x+h)) +sqrt(1+2x)

2/sqrt(1+2(x+h)) +sqrt(1+2x)


and Voila 1/sqrt(1+2x)

Edit. Left out a parenthesis

I know that's the right method, now, but I still am getting the wrong answer. When I do All the math I get 0 / (1+sqrt(1+2x)) instead of 1/sqrt(1+2x)....

 

[TuxDave]

check your work again? it looks like that posted solution works.... remember h/h = 1, not zero..heh

 

[VBboy]

LOL, sorry, I don't want to do it using the definition of the derivative. Here is the regular solution.

g(x) = sqrt(1+2x)

Which is the same as:

g(x) = (1+2x)^(1/2)
g'(x) = (1+2x)' * (1+2x) ^ (1/2 - 1)
g'(x) = (2) * (1+2x) ^ (-1/2)
g'(x) = -1/sqrt(1+2x)

I used to be George W's math advisor back in college, so I should know

 

[Woodchuck2000]

Originally posted by: VBboy
LOL, sorry, I don't want to do it using the definition of the derivative. Here is the regular solution.

g(x) = sqrt(1+2x)

Which is the same as:

g(x) = (1+2x)^(1/2)
g'(x) = (1+2x)' * (1+2x) ^ (1/2 - 1)
g'(x) = (2) * (1+2x) ^ (-1/2)
g'(x) = -1/sqrt(1+2x)

I used to be George W's math advisor back in college, so I should know

You've cocked up the last line - it should be g'(x) = 2/sqrt(1+2x)

It's possible to solve using a substitution as follows:

g(x) = sqrt(1+2x)

Let t=1+2x such that dt/dx = 2 and g(x) = sqrt(t)

now g'(x) = dt/dx * d(g(x))/dt where d(g(x))/dt = 1/sqrt(t)

g'(x) = 2/sqrt(1+2x)

 

[Ausm]

Originally posted by: amdskip
notfred mode <ON>: do your own homework for once so you actually learn something </OFF>I took calc 2 but that was too long ago. Sorry I can't help you.

Same here wait untill you take Calc 4 and Diff EQ


Ausm