Up for a calculus problem challenge? (YOU KNOW YOU WANT TO)

Test Preparation
1

<p>Okay, you gotta do these the long way using the definition of the derivative to find f'(x) (no shortcuts like the AP way)</p>

<p>f(x) = 3x + (6 / x²)</p>

<p>and</p>

<p>f(x) = ³root of x</p>

<p>I got stumped on both and didn't know where to move on. GOGOGO THX!</p>

<p>ITS LONG and I GOT STUCK AND IDK WHERE I WENT WRONG :(</p>

2

<p>okay dude when you post **** like that...i'm bound to look...now i'm blinded - i edited it so reread the post! find f'(x)</p>

3

<p>Okay, you know when you say Do Not Visit I'm going to click right? Do some people enjoy being reported? :confused:</p>

4

<p>definition of the derivative is:</p>

<p>f\'(x) = lim h->0 [f(x+h) - f(x)]/h</p>

5

<p>For the first one, it's 3 - 12/x^4.</p>

<p>Start doing the algebra, using teh (f(x+h)-f(x))/h lim h->0 derivative definitoni.</p>

<p>Get common denominators for the two fractions you'll get.</p>

<p>In the end, you won't get all the h's to cancel out, but that's ok. YOu only need the h in the denominator to go away, which it does. </p>

<p>Then, plug in 0 for h, and you'll get it.</p>

6

<p>yeah I know, and I did all the work and all, but got stuck and dunno where I messed up so if someone can be awesome and show all the work out on here!#!#!</p>

<p>Mr. Layman...can you show me the first few steps for the 2nd problem? I don't know how to manipulate cube roots...since u can't rationalize</p>

7

<p>Okay, for the second one, multiply numerator and denominator by:</p>

<p>x^(2/3) + x^(1/3)*(x+h)^(1/3) + (x+h)^(2/3).</p>

<p>That is the difference of cubes.</p>

<p>After that, the numerator should simplify out to h.</p>

<p>yeah, for the first one, just organize your algebra. My work is too too long to type out, sorry.</p>

8

<p>btw, you should get 1/3 * x^(-2/3) for the second one.</p>

9

<p>errr I'm so confused...for that second one I got up to this:</p>

<p>(x+h)^(1/3) - (x)^(1/3) / h</p>

<p>Idk what to do now</p>

10

<p>Now multiply the numerator and denominator of that fraction up there by the big expression i typed out 3 posts ago.</p>

11

<p>Okay, but what is the reasoning for doing this?</p>

<p>Also, when you multiply the top by that what should be the result?</p>

12

<p>I love conjugates.</p>