Find all relative maxima or minima.  y= (ln x)^2 , x>0?
399858
Correct, derive and solve for x=0, that will give you the maxima or minima.\nUsing product rule:\ndy/dx= 2ln(x)*(1/x).\nSolve for 2ln(x)/x=0; *x both side\n2ln(x)=0; /2 both sides\nln(x)=0; \nThe answer is 1. The natural log of 1 is zero.\nx=1\n\nTherefore, the min or max is at 1. \n\nIf you want to know if its a max or min you take the second derivative.\n\nd^2y/dx^2= 2/x^2-2ln(x)/x^2=2-0=2 -> If evaluated at the min/max value of x=1.\n\nSince the second derivative is positive, this function is convex, and at 1 it is minimized.
590831
y = x^2 - (54/x) = x^2 - 54x^(-1)\ndifferential of y w.r.t. x = 2x - 54x^(-2)\n\nWhen a line is perpendicular to the x-axis, it is parallel to the y-axis. So the gradient is underfined.\n\n2x - (54/x^2} is undefined when x=0\n\nSo the equation is x=0.
826336
y=6/ln(x)-9\nput x=e^9,y=infinity\nso x=e^9 is vertical asymptote\nlnx=(6/y)+9\nx=e^[(6/y)+9]\nx is not equal to infinity for any value of y...\nso no horizontal asymptote\n\n\nhope that helps
758848
let y = x^x\nw = ln y = xlnx\nthen y = e^w\ndy/dx = e^w * dw/dx\ndw/dx = x(1/x) + lnx = 1+lnx\ndy/dx = (1+lnx)*x^x
832985
Let S = x + 1/x.\n\nWe want to minimize x for x>0.\n\nTake the derivative of S with respect to x and set that equal to zero to find the extrema of S:\n\n0 = dS/dx = 1 - 1/x^2\n\n1/x^2 = 1\n\n1 = x^2\nx = 1 (and -1, but we are only interested in x>0 here).\n\nWe can check to make sure this is a minimum by looking at the sign of the second derivative of S with respect to x at x=1:\n\nd2S/dx2 = 2/x^3\n\nWhen x = 1, the second derivative is equal to 2, which is positive, so the extremum at this point is a minimum (the function is concave upward).
766935
The easy way is to note that both numbers have to be equal so 56/2=28\n\n\nthe hard calculus way is to maximize xy given that y=56-x\nx+y=56 implies\nxy=x(56-x)=56x-x^2= to find maximum of 56x-x^2\ndifferentiating with respect to x gives 56-2x. The critical point is when  56-2x=0 so x=28.y=28
792110
Thus y’=3x^2-6x+9=0 to find extrema, or x^2-2x+3=0, but (x-1)^2+2>0 - no extrema!
333624
(ln x)^7=(7*(lnx)^6)*(1/x)=7*(lnx)^6/x
740147
If you have a TI-83 you can graph it and find it easily...\n\ny= button (plug in equation) ---> hit graph\n\nthen hit the second button(yellow)--->calc(same as trace button)\n\nbecause the graph was concave up you want to find a minimum...so click that...and simply enter once on the left side of the graph, once on the right, and then guess in the middle.  It will then give you the exact answer.\n\n(1,6) should be the correct coordinate.
909799
gopal's solution:\n\nf(x)=x^3-6x^2+9x+2\nf'(x)=3x^2-12x+9=0 for extremas\n3(x^2-4x+3)=0\n3(x-3)(x-1)=0\nx=3 or 1\nf''(x)=6x-12\nat x=1 6(1)-12=-ve so max\nat x=3 6(3)-12=+ve so min\nso the absolute maximum is at x=1 and\n=6\nabsolute min is at x=3 and is=2  <------ However still need to check endpoints of the interval [-1,4] as these turnng points are actually LOCAL maxima and minima!\n\nSo f(-1) = -1 - 6 - 9 + 2 = -14 < 2\nSo ABSOLUTE minimum = -14 at x = -1\n\nAlso f(4) = 64 - 96 + 36 + 2\n= 6\nSo ABSOLUTE maximum = 6 and occurs at both x = 1 AND at x = 4
797160
Your function y=4*x^2*e^(5x) is defined for any x. Meanwhile x^2>=0 & e^(5x)>0, thus y>=0 for any x. Now y’=4*(2x+5x^2)*e^(5x), local extrema being at y’=0. Or x1=0, x2=-0.4\nHence and “above” x1=0 means local min & y(0)=0, so x2=-0.4 means local max, \nLim[for x->-infinity]=0 & Lim[for x->+infinity]=+infinity. Does it help you Xu?
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
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
971776
y=10^x = e^(x ln 10)\nApplying the chain rule:\ndy/dx = e^(x ln 10) * ln 10 = 10^x ln 10\nAnd in general:\nd(a^x)/dx = a^x ln a
133024
y = 0.\n\nETA: That's if we're to assume that the value of x is constant between the equations.
1060074
d (ln(x)) / dx = 1/x\n\nThe derivative of the natural log of x with respect to x equals 1 / x
1138866
I'll lead you in the most important part, but it is up to you to complete it.\n\nDerivative of x . y = x'.y + x.y'\n\nThen, you will have:\n\nf'(x) = (2x . ln x^2) + x^2 (ln x^2)'\n\nContinue from here, the only problem that you may have is finding the derivative of ln y , which you can see in the link below.