<p>adidasty...actually...the second one is a different answer than the one you proposed.</p>
<p>f(x)=xsin(ln5x)</p>
<p>f'(x)= sin(ln5x) + (cos(ln5x))/5</p>
<p>You forgot to do chain rule on the ln5x for the second term.</p>
<p>Psh...these problems are easy:</p>
<ol>
<li>f(x)= ln(ln(ln x)) - Answer: </li>
<li>w(x0= xsin(ln 5x) - Answer: </li>
<li>H(t)= lnt^2(3t^2+6) - Answer: </li>
<li>G(t)= ln{sqrt(5t+1)}( t^3+4)^6 - Answer:</li>
<li>f(x0= ln{(x+1)(x+2)}/x+3 - Answer:</li>
<li>f(x)= ln sqrt{(3x+2)^5/x^4+7} - Answer:</li>
</ol>
<p>I used the combined rule of derivatives plus some special font to post the answers.
The answers will only show up to true geniuses, you must figure out how to make them appear.
P.S. highlighting won't work. :)</p>
<p>Only a true genius can solve this: "Prove that this post does not exist."
Have fun.</p>
<p>Haha clever!!!:) Seriously, I am having a lot of trouble finding the derivative of these fuctions. Any help would be greatly appreciated!!!</p>
<p>One word. Mathematica (yes I know it's cheating. Don't use it if it's homework)</p>
<p>
[quote]
"Prove that this post does not exist."
[/quote]
To exist is to be existent; to be existent is to be actual; to be actual is to be real; to be real is to be genuine; to be genuine is to be true; to be true is to be honest; to be honest it to be guileless; to be guileless is to be transparent; to be transparent is to be clear; to be clear is to be unclouded; to be unclouded is to be light; to be light is to be insufficient; to be insufficient is to be inadequate; to be inadequate is to be lacking; to be lacking is to be nonexistent.</p>
<p>Therefore that post does not exist.</p>
<p>(courtesy of hyperdictionary.com)</p>
<p>jaug1- are you sure? heres my reasoning:</p>
<p>xsin(ln5x) d/dx = sin(ln5x) + x((sin(ln5x)) d/dx) =</p>
<p>y=sin u, dy/du = cosu</p>
<p>u = ln5x, du/dx = 5/5x = 1/x. dy/du * du/dx = cos(ln5x)/x (then you multiply by x and they cancel) =</p>
<p>d/dx (xsin(ln5x) = sin(ln5x) + cos(ln5x)</p>
<p>so yeah i got it wrong the first time, but i got it right this time. lol and YOU got it wrong! loser! lol jk</p>
<p>if i got this wrong **** it, i give up. im on drugs anyway. lol</p>
<p>keta, if I solve the hodge conjecture, Ill be claiming the money, so dont expect me to send you my solution :)</p>
<p>is the conjecture already proven?</p>
<p>tetrahedr0n, perhaps you can then share your solution at art of problem solving forums. You're registered there, right? ;).</p>
<p>tetrahedr0n, please tell me how to prove it..</p>
<p>i'm glad we did this several days after the question haha</p>
<p>Prove the conjecture please</p>
<p>I guess the title of this thread used to attract self-entitled geniuses, like the fire attracts heat-seeking insects.</p>
<p>
[QUOTE=chanman]
lol what the heck is this??</p>
<p>"On a projective non singular algebra variety over C, any Hodge class is a rational linear combination of classes cl(Z) of algebraic cycles. "
[/quote]
</p>
<p>chanman, this is what pure math looks like. When you’re in college, and your professors aren’t busy lecturing, this is the kind of thing they work on. And if any of us solve it, you’ll be more likely to hear about it in the newspaper than here.</p>
<p>Congrats you guys bumped a 4 year old irrelevant homework problem,</p>
<p>Haha. It came up on “similar threads” when I was posting, and I didn’t realize it was so old. Sorry.</p>
<p>Oops. Note to self: check thread date, even if recently posted to.</p>