y = x^lnx (please help me)?
347784
Taking natural logarithm to both sides\nLn(y)= Ln(x) * Ln(x) = (Ln(x))^2\nDifferentiating\n(1/y )dy/dx = (2/x) * (Ln(x))\ndy/dx = (2y/x) * (Ln(x))\nat the point (e,e),\ndy/dx = (2e /e) * Ln(e)\nso the slope = 2
785428
y=e^cos(x) + cos(e^x)\ny'=-sin(x)e^cos(x) + -sin(e^x)e^x\n=-(sin(x)e^cos(x)+sin(e^x)e^x)
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/
767881
ln( a^ b) = b ln a\n\nusing the above ln(y^52) = 52 ln y
803558
1.\nc=(xy^2-1)/y\ndifrentiate w.r.t x\n(y(y^2 + 2xyp)-p(xy^2-1))/y^2=0\ny^3 + 2xy^2p - pxy^2 + p=0\ny^3 +xy^2p +p=0\nwhere p=(dy/dx)\n2.\nsimilarly diffrentiate \n(x+y^2)/x^2=-c\nto get\n2(x^2)yp - x^2 - 2xy^2=0\nthanks
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
101438
Y=MX+B\nY=-X ------> Y=-1X+0 \nSLOPE IS -1 AND THE Y-INTERCEPT IS 0
974331
= ln x / x^5\nln x -> inf\nx^5 -> inf\ndifferentiate both sides (Lagrange)\n\n=> 1/x  / 5x^4\n= 1/ (5x^5)\n=> 0 \n\nOr ln x increases much slower as x gets bigger, whereas x^5 increases faster, (exponentially) as it gets bigger. polynomial wins over logarithm. \n\nThe limit is 0.
538185
Let u = ln(x+1) and dv = x^2dx, then\ndu = (1/(x+1))dx and v = (1/3)x^3\nuv- int(vdu)\n= (1/3)x^3ln(x+1) - (1/3)int(x^3/(x+1)dx)\nDivide x^3/(x+1) using long or synthetic division to get\nx^2 - x + 1 - 1/(x+1)\nint(x^2 - x + 1 - 1/(x+1))\n= (1/3)x^3 - (1/2)x^2 + x - ln(x+1)\n= (1/3)x^3ln(x+1) - (1/3)((1/3)x^3 - (1/2)x^2 + x - ln(x+1))\n= (1/3)x^3ln(x+1) - (1/9)x^3 + (1/6)x^2 - (1/3)x + \n(1/3)ln(x+1)
51150
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )
504123
Divide by 100^y in the first to get (1+x/100)^y=2. Now take natural logarithms of both sides to get\ny*ln(1+x/100)=ln(2).\nNow, it turns out that ln(1+x/100) is about x/100 for small values of x. This gives\ny*x/100=ln(2) or\ny*x=ln(2)*100.\nNow, ln(2) is about .693, so 100*ln(2) is about 69.3.\n\nThis just says that the rule of 72 should be the rule of 70, or even the rule of 69. But, 72 has a lot of factors, so it's easier to get approximate doubling times in your head by using 72.
787158
y = xtan(x)\n\nUse product rule:\ndy/dx = x/cos²(x) + tan(x)\n= x/cos²(x) + sin(x)/cos(x) \n= [x + sin(x)·cos(x)] / cos²(x)\n\nUse product and chain rules:\nd²y/dx² = 1/cos²(x) + (-2)·(x/[cos(x)]^3)·(-sin(x)) + 1/cos²(x)\n= 2/cos²(x) + [2·x·sin(x)] / [cos(x)]^3\n= [2·cos(x) + 2·x·sin(x)] / [cos(x)]^3
238752
ln and e^x are inverse functions. So you can use them to solve equations.\n\nA looks like it's already solved, so just use a calculator and find the ln of 0.0093, which is -4.67774\n\nFor equation B, take the ln of both sides.\n\nln(e^x) = ln(62.5)\n\nThe ln(e^x) cancels each other out.\n\nx = ln(62.5)\n\nx = 4.13517
817379
y=(x^2-3x)^2\ny=x^4-9x^2\nAnd then I'm confused.
275631
One way to see this is to write x^(1-x) as exp((1-x)ln x) and use the chain rule. Only one person has the correct answer, art_colls. The power rule *does not apply* since the exponent is a function also!
645941
4x + y >= 4\ny >= -4x +4\n\n-------\ny = 0\n0 >= -4x + 4\nx = 1\n-------\nx = 0 \ny >= 0 + 4\ny = 4\n-------\n\n\nTherefore, shade the region above the line.\n\nTwo points on the line are:\nx-intercept: (0, 4)\ny-intercept: (1, 0)