<li>f(x)= ln(ln(ln x))</li>
<li>w(x0= xsin(ln 5x)</li>
<li>H(t)= lnt^2(3t^2+6)</li>
<li>G(t)= ln{sqrt(5t+1)}( t^3+4)^6</li>
<li>f(x0= ln{(x+1)(x+2)}/x+3</li>
<li>f(x)= ln sqrt{(3x+2)^5/x^4+7}</li>
</ol>
<ol>
<li>f(x)= ln(ln(ln x))</li>
</ol>
<p>I'll get you started. Its actually not hard at all. Just remember to work from the outside and never touch the insides....chain rule</p>
<p>ok so you start with your outside function, differentiate knowing that the derivated for any function ln(x) will be 1/x</p>
<pre><code> 1
</code></pre>
<p>________ * f'(ln(ln x)
ln(ln x)</p>
<pre><code> 1 1 1
____ * _____ * _____
</code></pre>
<p>ln(ln x) ln(x) x</p>
<pre><code> 1
x* ln(x) * ln(ln x)
</code></pre>
<p>that's your answer</p>
<p>These really aren't hard, but more of a tedious excercise in making sure you use chain/product/quotient rules correctly. If you know those you should be fine.</p>
<p>That said, what are these problems for? They seem excessive to the point of futility for a calculus textbook, and I doubt you will see anything so long on the AP exam as a MC question.</p>
<p>Apparently "genius" is now defined as someone who knows two or three extremely basic formulas. Those exercises do not require any intelligence whatsoever. The last two are a bit tedious, but that is only a test of patience. What is the point of this thread?</p>
<p>I agree, nowadays people ascribe "genius" qualities both to themselves and others so often that it's getting to be very annoying and semi-idiotic.</p>
<p>Geniuses come along once or twice in a generation. </p>
<p>So get used to it, YOU'RE NOT THAT SPECIAL!!!</p>
<p>i think the point is getting someone to do his homework for him</p>
<p>this is easy../.</p>
<p>Then show me that you can do it...</p>
<p>come on... man on fire.</p>
<p>These problems only require that you know how to use the chain rule, product/quotient rule and that you know ur derivative rules. Are you in AB Calc? Because I am, and we finished this stuff in like October/November. I showed you how to do the first one. Just review your rules and you will see how easy they are.</p>
<p>c'mon just like these guys said- theyre chain rule and product rule things. like look:
f(x) = xsin(ln5x)</p>
<p>Product rule:</p>
<p>f'(x) = sin(ln5x) + xcos(1/x)</p>
<p>(because derivative of ln 5x = 5/5x = 1/x and the derivative of sin is cos).</p>
<p>man! we are doing this guy's homework!</p>
<p>Everyone's a genius in his own ways.... ;)</p>
<p>Oh BTW, here is a question only geniuses can solve: </p>
<p>Prove the following conjecture:</p>
<ol>
<li>On a projective non singular algebra variety over C, any Hodge class is a rational linear combination of classes cl(Z) of algebraic cycles. </li>
</ol>
<p>Have fun and take your time!</p>
<p>keta- okay, we'll start by assuming that any Hodge class is NOT a rational linear combination of classes cl(Z) of algebraic cycles....lmao oh the ways of math...</p>
<p>lol im glad that we've just helped man on fire get his homework done!</p>
<p>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. "</p>
<p>These problems are neither hard nor easy. By hard, I mean they are not tricky. They are merely brutal exercises in carefulness (viz. the chain rule). But, because of the brutality, they are not easy.</p>
<p>But, I concede - I can't do these easily. I do need some paper to work these.</p>
<p>fabrizio, its okay, not everyone can do super mental psycho math like me. lol just kidding man. yeah now that i think of it theyre pretty much straight exercises. not that hard but not that easy either. just carefulness is needed (wow wonderful diction on my part. lol)</p>
<p>These are somety tricky problems, so I can understand if you guys can't do them!!! Even my Calculus teacher found them to be quite puzzling!!! With the exception of #1, they were all given a difficulty rating of 5 on my AP Calculus Prep Review Book. Oh well, I guess I will just keep trying to crack these problems.</p>
<p>Prove the following conjecture:</p>
<ol>
<li>On a projective non singular algebra variety over C, any Hodge class is a rational linear combination of classes cl(Z) of algebraic cycles. </li>
</ol>
<p>^^, I'm going to do that. i'll get the answer to u</p>