The only formula that I come up with myself is the following:\n \nDirectional derivative=( Nobla f(x,y).unit vector or the u vector )/ [[ u ]]
121368
If the unit vector has components "a" and "b" (the unknowns for which you want to solve) then your first equation is\n\n1.  a^2 + b^2 = 1\n\nIf you have f and the directional derivative then you can take the partial derivatives of f with respect to x and y (call these f_x and f_y) and get your second equation\n\n2.  a f_x + b f_y = directional derivative\n\nWith two equations in the two unknowns a and b you then try to solve for a and b.
605584
differentiate both sides:\ncos(x^2/y^2) * derivative(x^2/y^2) = 1\ncos(x^2/y^2) * (y^2 * 2*x -  x^2*2*y*(dy/dx))/y^4 = 1\ncos(x^2/y^2) * 2xy(y-x*(dy/dx))/y^4 = 1\ny-x*(dy/dx) = y^4/(2*x*y*cos(x^2/y^2))\n(y - y^4/(2*x*y*cos(x^2/y^2)))/x = dy/dx\nAssuming I didn't make any errors.
497666
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
552905
Rewriting:\ny = x(16-x^2)^1/2\ndy/dx = [(16-x^2)^1/2]d/dx(x) + xd/dx[(16-x^2)^1/2]    chain rule\ny' = [(16-x^2)^1/2](1) + x[(1/2)(16-x^2)^(-1/2)](-2x)\n   = sqr(16-x^2) - 2(x^2)/sqr(16-x^2)\nTry the second derivative using the same technique for the argument of the sqr.  You can do it, just take as many steps as you need to keep the numbers straight.
225698
You can refer this book:\n\nDifferential equations : linear, nonlinear, ordinary, parial - A.C. King, J. Billingham, and S.R. Otto.
481967
You can of course use a brute force approach to get the derivative but a little finesse often can simplify things.  You have:\n\ny=(sin(x)/1-cos(x))^2 = (sin(x)^2)/(1-cos(x))^2\n\nSince sin(x)^2 = 1 - cos(x)^2 it follows that:\n\ny = (1 - cos(x)^2)/(1 - cos(x))^2\n\nBy doing a little algebra up fornt, the final result is quite a bit simpler than some of the others given.\nFactoring the numerator:\n\ny = (1 - cos(x))*(1 + cos(x))/(1- cos(x))^2 = (1 + cos(x))/(1 - cos(x)) \n\nNow your function is of the form\n\ny = (1 + u)/(1 - u) where u = cos(x)\n\ndy/dx = dy/du * du/dx = (1/(1 - u) + (1 + u)/(1 - u)^2)(-sin(x))\n\ndy/dx = -2*sin(x)/(1 - u)^2 = -2*sin(x)/(1 - cos(x))^2\n\nBy doing a little algebra up front, the final result is quite a bit sinpler than those above.
572920
Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
537675
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
387955
Using implicit differentiation, you get:\n\n-csc^2(y)(dy/dx) = 1 - dy/dx\n-csc^2(y)(dy/dx) + dy/dx = 1\n(dy/dx)[1 - csc^2(y)] = 1\n\nTherefore, dy/dx = 1/[1 - csc^2(y)]\n\nNow plug in your ordered pair to get dy/dx.  Remember csc (y) = 1/sin(y):\n\ndy/dx = 1/[1-1], which is undefined.  Therefore, you cannot write a tangent line for this ordered pair (there is a vertical asymptote at this point)\n\n------------\n\nHope this helps
352637
dy/dx=2xy\ndy/y=2xdx\nintegrate\nlny=x^2+C\nput in initial conditions\nln(1)=1+C\nC=0-1\n'e' each side\ny=e^(x^2-1)\n\ncheck in original equation\ndy/dx=2x*e^(x^2-1)\n2x*y=2x*e^(x^2-1)\nthis answer is correct\nat x=1.6\ny=e^(1.6^2-1)=4.758821245\n\nto apply euler's method\nto the initial value problem\ndy/dx=f(x,y),  y(x0)=y0\nproceed as follows\n(a)take x0 and Y0=y0 as starting\nvalues,choose a step size h,\nand set r=0\n(b)set x(r+1)=r+h\n(c)calculate an approximation \nY(r+1) to(y(r+1)),using the\niteration formula \nY(r+1)=Yr+hf(xr,Yr)\n(d)if furtherapproximate values\nare required,increase r by 1,and\nreturn to sub-step (b)\n\ni also have details of the \nclassical forth order runge-\nkutta method,but will keep \nthat for a later date\n\ni hope that this helps
544894
By definition of the derivative\ng'(x) = lim h=>0 [g(x + h) - g(x)] /h =  lim h=>0 [3xh + h^2]/h = \nlim h=>0 [3x + h] = 3x
31567
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).
594639
start with   d = m / v        multiply both side by v; \nyou get     v * d = m        divide both sides by d; \nyou get     v = m / d
471875
You need to sum the components. Let Fr, Fl, Fu, and Fd be to right, left, upward, and downward directed forces, respectively. Then\n\nFx = Fr - Fl\n\nand\n\nFy = Fu - Fd.\n\nThe right and upward directions are taken as the positive direction so that the minus sign indicates that the forces (Fl and Fd) are directed to the left and downward.\n\nThe magnitude of the resulting force is then computed as\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the direction can be found from\n\ncos(q) = Fx/F\n\nUse some reasoning to be sure that the direction you compute is consistent with the orientation of the force components.
390993
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
446385
u= x\ndv=cosx dx\n\ndu =dx\nv= sinx\n\n\nintegral of xcosx dx = xsinx - integral of sinx dx\n= xsinx + cosx +C\n`
410151
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc