Calculus - Derivate using limit def.

y = sqrt(2x + 1)

Using the limit definition, I just keep getting left with 2, which isn't right. It should
1/sqrt(2x + 1)

Here's what I'm doing:
y' = lim x->a (sqrt(2x + 1) - sqrt(2a + 1))/(x - a)) //Limit definition
Multiply by conjugate now... Left with
(2x - 2a)/(x - a)
Factor out 2, the (x - a) terms cancel out. Left with just 2. 

Obviously something is going wrong here, and I'm gonna feel stupid when it's pointed out but I'm not seeing it right now -_-
(sqrt(2x + 2h + 1) - sqrt(2x+1) )/h
mult by reciprical

(= 2x+2h+1 - 2x+1)/(sqrt(2x+2h+1) + sqrt(2x+1))
factor h and cancel
[h(2)/[2(sqrt(2x+2h+1) + sqrt(2x+1))]
h goes to 0, cancel h, add sqrt(2x+1) and cancel 2

= 1/sqrt(2x+1)

Wow I forgot to multiply the denominator each time. I feel dumb
Oh man. Derivatives using the limit definition. Those are not fun. I loose myself in them all the time; don't feel dumb, lol.
I got too used to the nice derivative rules heh
I just don't think the limit definition is acceptable. Too much algebra to screw up.
Just wait until you review it in multi-variable.
I've got a couple semesters until that. I don't believe we get into multi-variable calculus until the third semester.
Same here, I'm in calc 3 now.
Ah nice how is it? I think working with 3D shapes sounds interesting.
I'ts surprisingly difficult to visualize functions in 3D, making graphing a pain, at least for me. The actual math isn't all that much harder, we are finishing partial derivatives and implicit differentiation which were fairly simple.

If you want to see some interesting stuff check out khan academy's videos on double and triple integrals, we haven't gotten to them yet but I watched the videos and found them very interesting.
We actually just talked about implicit differentiation in class today. I assume it's not much different with a third variable involved.

I'll have to check that video out, khan academy is good stuff.
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