Find dy/dx and d^2y/dx^2 at the point (2,1) on the curve x^2 + 2xy - y^2=7.?
820064
x^2+2xy-y^2=7\n\ndy/dx\n2x+2y+2x(dy/dx)-2(dy/dx)=0\n(2x-2)(dy/dx)=-2x-2y\n\ndy/dx= (-2x-2y)/(2x-2)\ndy/dx=-3\n\nd^2y/dx^2\n[(2x-2)(-2-2(dy/dx))-\n(-2x-2y)(2)]/(2x-2)^2\n\nd^2y/dx^2=5\n\ni think thats rite sry bout all the parantheses
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
968996
y^2-2xy-8=0...........(1)\nor, 2y(dy/dx) - 2 (x dy/dx + y) = 0\nor, y(dy/dx) - x(dy/dx) = y\nor, dy/dx = y/(y-x)\n\nLet us choose an arbitrary pt (1,k) on st line x=1\n\nhence, dy/dx(at (1,k) = k/(k-1).\nequation of the tangent on (1) at (1,k) is,\n\n(y-k) = k/(k-1) (x-1)\nor, (y-k)(k-1) = k (x-1)\nor, yk -k^2 -y +k = kx -k\nor, kx + y(1-k) = 2k - k^2................(2)\n\nfrom (1),\nputting x =1\ny^2-2y-8=0\nor(y-4)(y+2) = 0\nor, y = 4,-2\nhence k can assume the values 4 & -2\nputting the values of k in (2)\n4x + y(1-4) = 2*4 - 4^2\nie, 4x - 3y + 8 =0\n\n&\n-2x + y(1-(-2)) = 2*(-2) - (-2)^2\nie, 2x - 3y = 0\n\nHence the equations are 4x - 3y + 8 =0 & 2x - 3y = 0
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
364921
Solve for y\nSubtract 2x from each side\n-8y = -2x + 7\nDivide everything by -8\n-8y/(-8) = -2x/(-8) + 7/(-8)\ny = (1/4)x - 7/8\nFor the equation y = mx + b, the slope is m.\nAnswer: slope = 1/4
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.
360927
y'=dy/dx=\n\n(6√x)' + (3/x^2)' + (10)'=\n\n[6*x^(1/2)]' + [3x^(-2)]' + 0= \nnow use the power rule\n\n6*(1/2)* x^[(1/2)-1] + 3*(-2)*x^(-2-1)=\n\n3*x^(-1/2)-6x^(-3)=\n\n3/x^(1/2) -6/x^3=\n\n3/√x - 6/x^3
931612
f^{-1}(y) = (y-42) / 7\ns
1051703
collect like terms on one side\nxdx/(1+x^2)=dy/[(1+y^2)*y]\n\n1/2* 2x/(1+x^2)= dy/y-1/2*2y/(1+y^2)\n\nBy integrating each side,\n1/2*ln(1+x^2)\n=ln|y|-1/2*ln(1+y^2)\n\nuse log rules, take square of both sides,\n\nC*y^2 / (1+y^2)=1+x^2\n\ny=sqrt([1+x^2]/[C-1-x^2])
632411
yes u mean like the turning point of a parabolic curve...yea the x-coordinate is equal to -b/2a\nor u can differentiate the equation of the curve and put it equal to zero and solve :\neg  y = ax^2 + bx + c\ndy/dx = 2ax +b \nput this equal to zero and u get x = -b/2a
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
697906
let y = arc tan (x)\n\ntan y = x\ndifferentiate\n\nsec^2 y dy/dx = 1\n\nor(1+tan ^2y) dy/dx = 1\n\ndy/dx = 1/(1+tan^2 y) = 1/(1+x^2)
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)
1027274
2x^2 - 2y^2 = 2(x^2 -y^2) = 2(x-y)(x+y)\n-2x^2 + 2xy = -2x(x-y)\nx^2 + xy = x(x+y)\n2(x-y)(x+y)/[-2x(x -y)]/[x(x+y)]\n= 2(x-y)(x+y)/[-2x(x-y)(x)(x+y)]\n= -1/x^2
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)).