Find dy/dx by implicit differentiation if y = ln(xtany).?
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
732220
dx/dt=1/2\ndy/dt=2t\ndy/dx=dy/dt/dx/dt\n=2t/1/2=4t\nat t=-1 slope=-4\nat t=1 slope=4
982897
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
904114
First, put your original equation as Y1\n\nFor Y2, type  nDeriv(Y1, x, x)\n\npress the buttons   (math)(8)   to get   nDeriv\nthe buttons (VARS)(1)(1)  to get Y1\n\n\nFor Y3, type  nDeriv(Y2, x, x)\n\nthen just display graph Y3\n\nIt will take quite a long while to graph, better if you manually differentiate the original equation once, then use the nDeriv to do second derivative.
986332
y=ln[sec5x+tan5x]\ny'={1/[sec5x+tan5x]}[5sec5xtan5x+5sec^25x]\n=5sec5x[sec5x+tan5x]/[sec5x+tan5x]\n=5sec5x
806346
using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know
703729
You want to start by integrating the equation. Fortunately, this is not difficult.\n\ndy/dx = (x-2)e^(-2y)\n\nInt(e^(2y)dy) = Int((x-2)dx)\n\n (1/2)e^(2y) = (1/2)x^2 - 2x + C, where C is some constant\n\ne^(2y) = x^2 - 4x + C  Note that e^(-2y) = 1/[x^2 - 4x + C]\n\n2y = ln (x^2 - 4x + C)\n\ny = (1/2)ln (x^2 - 4x + C)\n\nWe are almost done, now we just need to plug in the given condition, and solve for C.\n\nln (2) = (1/2)ln(4 - 8 + C)\n2ln(2) = ln(C-4)\n4 = C - 4\nC = 8\n\nSo the particular solution is y = (1/2)ln (x^2 - 4x + 8)\n\nIf you want to check this, we need to differentiate it\n\ndy/dx = (1/2)(2x-4)/(x^2 - 4x + 8)\n\nRemember from above that  e^(-2y) = 1/[x^2 - 4x + 8]\n\nSo dy/dx = (x-2)e^(-2y) as required
1055222
Is that e^y?\n\nCompute the differential of both sides\n\n7dxe^y + 7xe^ydy + 9x^2dx = 20y^3dy\n(20y^3 - 7xe^y)dy = (7e^y + 9x^2)dx\ndy/dx = (7e^y + 9x^2)/(20y^3 - 7xe^y)
120944
dy/dx = -4x \n\nAt x=2, gradient is -8\n\ny - y(1) = dy/dx(x - x(1)) [x(1), y(1) means x,y coordinates]\n\ny - 1 = -8*(x - 2)\n\ny = -8x + 17
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
830737
I do not understand what it is you are trying to solve. Are yoiu trying to find the inverse fuction? If you are that will be extremely difficult. If you seek the derivative then it has been doone above for you.\n\nIf you seek the integral then it would probably requiire a power series But as it is your problem is not a problem as you have not stated what you want solved.\n\ny = x^x\n= e^xlnx\n\ndy/dx = (lnx + 1)x*x\n= (lnx +1)*y\nSo lnx = 1/y *dy/dx - 1\nie x = 1/e * e^(1/lny * dy/dx)\nWhence x = 1/e * e^ [1/(lny* dx/dy)]\n\nNow I find that unconvincing as to get dx/dy I need to know x in terms of y Its looped onto itself!!!! In fact it is a differential equation not a function. ( x = 1/e * e^ [1/(x' . lny)]
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
1081914
(dx/dy) = (xy^3) / (2y^4 + x^4)\n\nLet x = y*u.\nThen (dx/dy) = u + y*(du/dy).\n\nSubstituting these in the given equation we have,\n\nu + y*(du/dy) = (u*y^4) / {2y^4 + (u*y)^4} = u/(2 + u^4)\n\nor y*(du/dy) = u/(2 + u^4) - u = -u*(1 + u^4)/(2 + u^4)\n\nor ∫ [-(2 + u^4)/{u*(1 + u^4)}] du = ∫ [1/y] dy _________(1)\n\n\n\nNow, let us first express [-(2 + u^4)/{u*(1 + u^4)}] in easily integrable form.\n\n-(2 + u^4)/{u*(1 + u^4)} \n\n= -{2*(1 + u^4) - u^4}/{u*(1 + u^4)}\n\n= -{2*(1 + u^4)/{u*(1 + u^4)} + u^4}/{u*(1 + u^4)}\n\n= -2/u + u^3/(1 + u^4)\n\nTherefore, ∫ [-(2 + u^4)/{u*(1 + u^4)}] du\n\n= ∫ [ -2/u + u^3/(1 + u^4)] du\n\n= -2∫ [1/u] du + ∫ [ u^3/(1 + u^4)] du \n\n= -2*ln(u) + (1/4)*ln(1 + u^4) + k\n\nwhere k = arbitrary constant of integration\n\nHence, from (1) we have \n\n-2*ln(u) + (1/4)*ln(1 + u^4) + k = ln(y)\n\nor  (1/4)*ln(1 + u^4) + k = ln(y) + 2*ln(u) = ln(y) + ln(u^2) = ln(y*u^2)\n\nor k = ln(y*u^2) - (1/4)*ln(1 + u^4) \n\n= ln(y*u^2) - ln[(1 + u^4)^(1/4)]\n\n= ln[(y*u^2)/{(1 + u^4)^(1/4)}]\n\n\nNow, substituting u = x/y we have\n\nk = ln[(x^2/y)/{(1 + (x/y)^4)^(1/4)}]\n\nor k = ln[(x^2)/{(x^4 + y^4)^(1/4)}]\n\nLet k = ln(c).\n\nThen c = (x^2)/{(x^4 + y^4)^(1/4)}\n\nor x^2 = c*(x^4 + y^4)^(1/4)
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
1034892
What is the function? I assume it is \n\n(dy/dx is equal to 1 only if y=x+C whre C is a constant)\n\nTo prove use the definition of the derivative which is dy/dx as a limit of delta_x appraching of (delta_y/delta_x)\nwe have \ny2-y1=x2+C-x1-C then\ndelta_y=x2-X1\n\ndelta_y/delta_x=(x2-x1)/(x2-x1)=1 QED\n(the is no need to take a limit, however in must be used. I hav eommited it for lack of better representation)\n\nBest
798496
Let y = lnu where u = lnv where v = ln(6x)\n\ndy/dx = dy/du . du/dv . dv/dx\n= 1/u . 1/v . 6/6x\n= 1/ln(ln(6x)) . 1/ln(6x) . 1/x\n= 1/[xln(6x)ln(ln(6x))]