<p>@Alexmitu,</p>
<p>Haha, go fly then 
</p>
<p>But recall that I never said I mastered it–I simply learned enough to pull out a 5 on the exam. There is a difference bro :D</p>
<p>But really, AP Calc ain’t even dat hard. Derivatives, Integrals, bada-boom bada-bing DONE. (heehee sorry for kinda trolling on this sentence, but I promise my story is true!)</p>
<p>Bioboy - funny post
And surprisingly enough, I’m enjoying it and not having much trouble. Derivatives are cool.</p>
<p>4x^2
Dx/dy = 8x^1 (right?)</p>
<p>Haha, well I’m making progress…</p>
<p>^ If that’s written the way I think it is then yes.</p>
<p>:D awesome. </p>
<p>If you want to solve a problem since you guys seem to like it…</p>
<p>Find f’(x) if f(x) = cos(sqrt 3x).</p>
<p>^^^ I agree with him</p>
<p>@tizzy26 You are wrong. 3cos(x)^2 is not equal to 3cos^2(x) or 3(cosx)^2</p>
<p>Let me explain how this problem is solved correctly using the product and power rules for differentiation.
First, to clarify, cosx^2=cos(x^2). (cosx)^2 is written as cos^2(x)
d 3cos(x^2)= (d 3)<em>(cos(x^2)+3</em>d cos(x^2) by product rule</p>
<p>d cos(x^2)=(d x^2)*(-sin(x^2))=-2xsin(x^2)</p>
<p>So then, (d 3)<em>(cos(x^2)+3</em>d cos(x^2) =0<em>(cos(x^2)+3</em>(-2xsin(x^2))=-6xsin(x^2)</p>
<p>@Excavalier. If we take the integral(anti-derivative) of what you got we get 3(cosx)^2+c, which as I wrote before is most commonly written as cos^2(x) (w/out the parenthesis. I just added them because otherwise it looks like cos to the power of 2x, which just does not make sense)</p>
<p>From a former Calculus student.</p>
<p>Now that one was pretty easy. Try this, take the indefinite integral of sin(x^2) dxGood luck, you’re gonna need it. :D</p>
<p>@nomad17 please don’t use ambiguous notation, I assume you mean sqrt(3x), but I can’t be sure.</p>
<p>@michael2 sorry I don’t know how to write it in… </p>
<p>It has parenthesis and inside the parenthesis there is a square root of 3x.</p>
<p>Oops I did it wrong</p>
<p>But the fact that I am doing Calculus at 1:14 in the morning… god I am such a nerd</p>
<p>Excav- Is your answer -3sin(sqrt3x)/2sqrt3x?</p>
<p>Sorry about the ambiguous notation again…</p>
<p>Find f’(x) if f(x) = cos(sqrt 3x). </p>
<p>Just change it to this</p>
<p>Find f’(x) if f(x) = cos(sqrt(3x)). OR Find f’(x) if f(x) = cos[sqrt(3x)].</p>
<p>For the denominator I had 2(x^(3/2)) but I’m too tired to check my work. I’ll get back to this after some sleep haha 
sorry</p>
<p>Haha cool, I’ll post some more when you come back if ya want haha.</p>
<p>You have to use the chain rule to solve this and you will find:
f’(x) = -3/((2x^(1/2))*sin[sqrt(3x)]</p>