Derivative of: y=(xsin(xe^x))/(cosh(x+x^2))?
971224
(Couldn't see the full function, it's truncated, but when you click Reply you can see: \ny=(xsin(xe^x))/(cosh(x+x^2))\n\nAnyway this is pretty nasty, the answer will use lots of the Chain rule, Product Rule and Quotient rule.\nd/dx (uv) = ( v du/dx - u dv/dx ) / (v^2)\nwith: u = xsin(xe^x), v = cosh(x+x^2)\n\nSo: du/dx = xsin(xe^x) (apply Product rule)\n = sin(xe^x) + x.cos(xe^x).(e^x+xe^x)\n = sin(xe^x) + x(x+1)e^x.cos(xe^x)\n\nAnd: dv/dx = sinh(x+x^2) . (1+2x)\n\nNow just plug em in and crank it out:\n\nd/dx (uv) = ( v du/dx - u dv/dx ) / (v^2)\n\n = [ cosh(x+x^2) * (sin(xe^x) + x(x+1)e^x.cos(xe^x))\n     - xsin(xe^x) * (sinh(x+x^2) * (1+2x)) ]\n/ ( cosh(x+x^2) )^2\n\n= [ cosh(x+x^2) * (sin(xe^x) + x(x+1)e^x.cos(xe^x))\n     - x(1+2x)sin(xe^x)sinh(x+x^2) ] / ( cosh(x+x^2) )^2\n\nExtracting a cosh(x+x^2) factor:\n\n= [ (sin(xe^x) + x(x+1)e^x.cos(xe^x))\n     - x(1+2x)sin(xe^x)tanh(x+x^2) ] / cosh(x+x^2)
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
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
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.
965203
y = (ln x²)^(2x+3) = e^((2x+3) ln ln x²)\nNow just ruthlessly apply the chain rule:\ndy/dx = e^((2x+3) ln ln x²) * (2 ln ln x² + (2x+3) * 1/ln x² * 1/x² * 2x)\nAnd simplify:\n(2 ln x)^(2x+3) * (2 ln (2 ln x) + (2x+3)/(x ln 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.
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
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.
786389
The derivative is [(x-4)- (x+1)]/(x-4)^2 = -5/(x-4)^2\n\nIf we use the definition of the derivative:\n\n\ndy/dx = Lim h -> 0 {f(x+h) -f(x)}/h\n= lim h-> 0 [(x+h+1)/(x+h-4) - (x+1)/(x-4)]/h\n                   (x+h+1)(x-4)-(x+1)(x+h-4)\n= lim x-> 0 ---------------------------------------\n                   (x+h-4)(x-4)h\n\n= lim h-> 0 -5h/(x+h-4)(x-4)h = -5/(x-4)^2
940999
Differentiate the function as you would normally except that when the derivative of y is taken make sure to replace y with y'. So this would be,\n(1/2)(xy)^(-2/3)(xy') + y = 0 - y'\nThen, solve for y' by moving all the y' over to one side and factoring it out.\ny' + (1/2)(xy)^(-2/3)(xy') + y = 0\n\ny'((1/2)(xy)^(-2/3)x + 1) = -y\n\ny' = -y / ((1/2(xy)^(-2/3)x + 1)\n\nThe second derivative is (quotient rule),\ny'' = (.5(xy)^(-2/3) + 1)(-y') - (-y)(1/3)(xy)^(-5/3)xy(xy')(y)\n\ny'' = .5(xy)^(-2/3)y' - y' + (1/3)x^2*y^3*y'(xy)^(-5/3)\nFactor,\ny'' = y'{(xy)^(-5/3)[.5 - 1 + (1/3)x^2*y^3(xy)]}\n\ny'' = y'{(xy)^(-5/3)[- .5 + (1/3)x^2*y^3(xy)]}\n\n(check my work)
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
980558
∫xe^x dx \n= ∫x d(e^x)/dx dx \n= xe^x - ∫e^x (dx/dx) dx\n= xe^x - ∫e^x dx\n= xe^x - e^x + c\n= (x - 1)e^x + c
139329
calculus, fun stuff. To get the derivative, you need to drop the exponent then subtract one from the explonent. SO the X^2 would be 2x and 2x would just be 2 .    Final answer is 2x + 2.   Hm.. explained that poorly. In general, to find the derivative, Take the exponent on the X, drop it in front so its multipluing, than subract one from the exponent and have that number in the old exponents place (3x^3 would turn into 9x^2, 4x^2 would be just 8x) any number with X^1 is just the number, and just a number with no X is 0.
797614
The only problem in your derivation is that you are evaluating f(0, 0) as 0/0. This is not the case, the second part of the definition is that f(x, y) = 0 when (x, y) = (0, 0). Nonetheless, your result is correct: f_x(0, 0) = [h→0]lim (0/h² - 0)/h = 0. Similarly with f_y(0, 0).\n\nThis makes sense, because the restriction of f(x, y) to the x-axis is a constant function (i.e. f(x, 0) = 0 for all x), so of course the partial derivative of it w.r.t. x is constant. Similarly, if you restrict f(x, y) to the y-axis, you also get a constant function.\n\nNote that f(x, y) is not differentiable in the 2-variable sense at (0, 0) (there is no tangent plane at that point), so the multivariate chain rule will NOT hold, and the directional derivatives won't be related to the partial derivatives. This helps to highlight an important difference between the partial and total derivates - it is possible for all partial derivatives to be defined and for the function to still not be differentiable at a point.
376765
Let u = 1 + x^4 and then apply the Chain Rule. This allows your expression to be written as\n\n1/sqrt(u)^(1/2)\n\nwhich is equivalent to\n\n1/u^(1/4)\n\nwhich, in turn, is equivalent to\n\nu^(-1/4).\n\nDifferentiating this expression with respect to x yields the following:\n\n(-1/4)*u^(-1/4-1)*(du/dx).\n\nNote: du/dx = 4x^3.\n\nTherefore, the desired derivative is \n\n(-1/4)*(1+x^4)^(-5/4)*(4x^3).\n\nIf we simplify this a bit we are given\n\n-x^3/(1+x^4)^(5/4).
948239
xy = y^½\n y = y^½ /x\n y = y^½*x^-½\n\ndy/dx = -½ y^½ x^ -3/2\ndy/dx = - [½ y^½ ]/x^+3/2\ndy/dx = - (y^½) / (2x^3/2)
845493
we shall use chain rule\n\nd/dx(sec(pi*x)^2\n= - sec(pi*x) csc(pi*x)d/dx(pi*x))^2\n= -2pi^2x sec(pi*x) cosec(pi*x)