<p>I'm having a hard time finding an antiderivative of sinxcosx. I think I found it, but I'm not sure. I used a u-substituion for sinx.</p>
<p>So,</p>
<p>let u= sinx
du=cosxdx
Thus, udu= (1/2)u^2= (1/2)sinx^2= ((sinx)^2)/2</p>
<p>Is that right? If it isn't, please correct me.</p>
<p>yeah, that's right.</p>
<p>All right. Thanks. I have a huge test tomorrow, and I think I understand this material.</p>
<p>Good luck. Are you in Calc AB Or BC?</p>
<p>I have a big chapter test tomorrow on that same material as well. (AB)</p>
<p>Me too lol.</p>
<p>I have one next week. Application of integrals.</p>
<p>I have one this Thursday :eek: I am so confused. No wonder why my name is dazed&confused :D Can you please help me? Thanks in advance!</p>
<p>1) derivative of x^(x-1)</p>
<p>2)derivative of (y+1)/(y-1)</p>
<p>3) e^(2x)/(1+(e^x))</p>
<p>4) Antiderivative of (lnx)x(4^xsquared)</p>
<p>Hmm...it sounds like your homework. Nonetheless, I'm going to try to solve those problems to help me review for the test.</p>
<p>1) y=x^(x-1)
lny= lnx^(x-1)
lny= (x-1)(lnx)
y'/y= [x*(1/x)] + lnx
y'/y= 1+lnx
y'= y(1+lnx)
y'= (x^(x-1))(l+lnx)
Then simplify when necessary</p>
<p>Please correct me if I'm wrong.</p>
<p>2.
(y+1)/(y-1)
((y-1)(y) - (y+1)(y)) / (y-1)^2
-2y / (y-1)^2</p>
<p>I understand the problem-solving processes for number 1 and 2. But, how can you find an antiderivative of number 2: (y+1)/(y-1)</p>
<p>Whoops. I'm wrong. </p>
<p>1) y=x^(x-1)
lny= lnx^(x-1)
lny= (x-1)(lnx)
y'/y= xlnx-lnx
Then find the derivatives of xlnx & lnx
Multiply it by x^(x-1)</p>
<p>Sorry</p>
<p>correction for hairypotty:
1) y=x^(x-1)
lny = (x-1)lnx
(1/y)(dy/dx) = (x-1)/x + lnx
dy/dx = y (lnx + (x-1)/x)
For number 2, I dont know if it asks dy/dx or d/dy. That will be 2 different problems then...
3) do you want the integral or deri. ???</p>
<p>and for number 4, are you sure it's not antideriv. of x(ln4)(4^xsq)??</p>
<p>2)d/dy
3)Anti.
4) Yes, anti.</p>
<p>that will be a lot easier then ^<em>^ lol, I'm a calc lover! ok, ready
2/ Quotion rule:
d/dy = [(y-1)1 - (y+1)1]/(y-1)^2
d/dy = -2 / (y-1)^2
I'm still working on the last 2, I learned it a long time ago, hard to remember it ></em><</p>
<p>That makes sense. How is d/dy different from dy/dx? How do you do it?</p>
<p>I have a calc ab test tomorow as well O__O</p>
<p>Today's calc study group was the best ever....</p>
<p>d/dy is the derivative of that function with respect to y. dy/dx is the derivative of function y with respect to x (in another word, the change in y with restpect to the change in x)
For the last two, I have no idea how to do it, honestly... I feel retarded... r u sure you copy it correctly from the textbook or wherever it is? 'cause I cant do it, and I'm in BC, what ashamed...!!</p>
<p>Hey guys! How did your calc test go? My test wasn't so bad, and surprisingly, the problem I tried to solve (but got an incorrect answer) was on the test! Thanks dazed&confused & van1011! I would lose a few points without you! :D</p>