y=e^x-e^(-x))/(e^x+e^(-x))? Find y'.?
108082
Using quotient rule, chain rule, and the knowledge of the derivatives of transcendtal functions(here the exponential function), you will arrive at:\n\ny'=1/(cosh(x))^2\n\nwhere cosh(x) is the hyperbolic cosine function - \n\ncosh(x) = (1/2)*(e^x+e^(-x))\n\nThe details of the work get kind of messy, which is why I left them out, but it is really just a standard quotient rule problem. Review your quotient rule, and you should have no problem.
631400
Factor out the y:\n\n(a-2) * y = c\n\nThen divide both sides by (a-2):\n\ny = c / (a-2)
364111
take the first derivative and set it equal to zero and solve.  2x +1 = 0 x= -1/2  solve for y.  y=-13/4\nvertex is at x= -13/4 y=-2
884999
The homogeneous part has solution y_h=Ce^(-x). For the particular solution, try y_p=Ax+B and find y_p=-x+1, so the complete solution is\ny=e^(-x) -x+1.\n\nAlternatively, use e^x as an integrating factor. When you multiply by this, the equation becomes becomes\nd/dx[e^x y]=-xe^x.\nNow integrate both sides and multiply through by e^(-x) to get the solution.
934968
Okay, assuming you mean:\n∫e^(2x)/√(1-e^(2x)) dx\nLet u=1-e^(2x), du=-2e^(2x) dx, dx = -1/2 e^(-2x)\n-1/2 ∫1/√u du\n-1/2 ∫u^(-1/2) du\n-u^(1/2) + C\n-√(1-e^(2x)) +C
979217
y = sqrt(1+t^2) = (1+t^2)^(1/2)\n\nby the power and chain rule\n\ndy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)\n\n= 2t(1/2) /sqrt(1+t^2)\n\n= t/sqrt(1+t^2)\n\nTherefore\n\ndy = (t / sqrt(1+t^2)) dt
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
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)
106013
y = -2x + 8\n\nsolve it yourself
270795
so y= (x^(7/17))*((x-7)^2)\n  y'=((7/17)x^(7/17-1)) ((2)(x-7)^(2-1)(1))\n  y'=(7/(17x^(10/17)) (2(x-7))\n  Y'=(14x^(7/17)) (49/17^(10/17))\nand then\nset y'=0\nsolve for x.....
847176
f(x,y) = x^2*y*(e^(-x^2-y^2))\n\nfx(x,y) = 2x*y*e^(-x^2 - y^2)(1-2x^2) this is wrong:\nf_x(x,y) = x^2y ( e^(-x^2-y^2)) (-2x) + 2xye^(-x^2-y^2))\n= 2xye^(-x^2-y^2) [-x^2 + 1] =0\nif x=0 or y=0 or x^2 =1, \nx=0, y=0, x=1 or x=-1\n\nfy(x,y) = x^2*e^(-x^2 - y^2)(1-2y^2) \nf_y(x,y)=  x^2 y (e^(-x^2-y^2)) (-2y) +  x^2(e^(-x^2-y^2))\n=x^2e^(-x^2-y^2) [-2y^2 + 1] \nx=0, y=1/sqrt(2),y=-1/sqrt(2)\n\nso the critical points are: (0,y)\n(0,0), \n(1,1/sqrt(2)), (1,-1/sqrt(2))\n(-1,1/sqrt(2)),(-1,-1/sqrt(2))\ns
169494
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
765615
differentiate implicitly:\n\n2x + 2yy' = 0\n\nand then some algebra\n\ny' = -(x/y)\n\nthen differentiate again:\n\ny'' = (-y - (-x)*y')/y^2 = (-y + xy')/y^2\n\nthen substitute in values:\n\ny' = -0/5 = 0\n\nand y'' = (-5 + 0*0)/(5^2) = =-5/25 = -1/5
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )
300751
Use the chain rule and the power rule.\n\nIf y = x^a, then y' = a * x^(a-1).\n\nSo, y = (2x+3)^(.5)\ny' = .5 * 2 * (2x+3)^(-.5) = (2x+3)^(-.5)\ny'' = -.5 * 2 * (2x+3)^-1.5 = -(2x+3)^(-1.5)
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
13683
Let e^x equal some variable (let's use y).\ny+1/y=4\nMultiply both sides by y.\ny^2+1=4y\ny^2-4y+1=0\nNow you can use the quadratic formula to find y.\ny=2+2√2 and 2-2√2\nBut don't forget that we said y=e^x, so make sure to solve for x.\nln(2+2√2)=x (In this case, y cannot be negative because one cannot take the natural log of a negative number)\nx≈1.175