Use implicit differentiate to find y', where 5xy/x+y = (x^2 + 2)?
957977
5xy=(x+y)(x^2+2)\n5xy=x^3+2x+x^2y+2y\n5xy'+5y=3x^2+2+x^2y'+2xy+2y'\n5xy'-x^2y'-2y'=3x^2-5y+2xy\ny'(5x-x^2-2)=3x^2-5y+2xy)\ny'=(3x^2-5y+2xy)/(5x-x^2-2)
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
199440
find the tangent of the equation x+y=5
152408
Use excel to give delta y/delta x on the curve. Look for a point where this value (the derivative) shoots up in value. That's your point.
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)
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.
739543
To differentiate y= 2/x\n\nFirstly you get rid of the fraction by taking X to the top(this will make its power negative) giving:\n\ny=2x^-1\n\nFrom here its just a simple case of carrying out the differentiation:\n\ndy/dx = -2X^-2 or dy/dx = -2/X^2
801587
Are you in pre-calc or calculus yet?  If so, you could find the vertex by taking the derivative of the function and finding where it equals zero (the derivative = the slope of the tangent to the curve, and the tangent to the curve = 0 at the vertex).\n\nf(x) = 5x² - 25x + 5\n\nf '(x) = 10x - 25\n\nf '(x) = 0 when x = 2.5\n\nSo, now find f(2.5) to determine the y-coordinate of your tangent:\n\nf(2.5) = 5(2.5)² - 25(2.5) + 5\n= 31.25 - 62.5 + 5\n= -26.25\n\nSo your vertex is at the point (2.5,-26.25)\n\n~ ♥ ~
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
795130
y = tan(πx - 1)\nlet u = πx - 1\nnow, dy/dx = (dy/du)(du/dx)\nd(tan(u)/du = sec^2(u)\ndu/dx = π, so \ndy/dx = πsec^2(u)\ndy/dx = πsec^2(πx - 1)
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
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)).
456921
assuming this is your question\n? / (5xy-9x) = (3+25y) / (9-5y)\n? = -x * (9-5y) * (3+25y) / (9-5y)\n? = -3x -25xy\n\nnow pay up
707539
x = (y^2 + 2y -1) / 2\ndx=ydy+dy\n1=(y + 1 )dy/dx\ndy/dx=1/(y+1)\nd2y/dx2 = -1 / (y+1)^2
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
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.
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )