Calculate dy/dx for Parametric Equations x(t) = 1/2 sin(2t^2) + t and y(t) = e^(2t) cos(t^2)

To\u0020find\u0020dy/dx,\u0020we\u0020need\u0020to\u0020differentiate\u0020y(t)\u0020with\u0020respect\u0020to\u0020x(t).\u000a\u000aFirst,\u0020let's\u0020find\u0020dx/dt.\u0020We\u0020have\u0020x(t)\u0020=\u00201/2\u0020sin(2t^2)\u0020+\u0020t.\u0020Taking\u0020the\u0020derivative\u0020with\u0020respect\u0020to\u0020t:\u000a\u000a dx/dt\u0020=\u0020d/dt\u0020(1/2\u0020sin(2t^2)\u0020+\u0020t)\u000a =\u0020d/dt\u0020(1/2\u0020sin(2t^2))\u0020+\u0020d/dt\u0020(t)\u000a =\u0020cos(2t^2)\u0020*\u0020d/dt\u0020(2t^2)\u0020+\u00201\u000a =\u0020cos(2t^2)\u0020*\u00204t\u0020+\u00201\u000a =\u00204t\u0020cos(2t^2)\u0020+\u00201\u000a\u000aNow,\u0020let's\u0020find\u0020dy/dt.\u0020We\u0020have\u0020y(t)\u0020=\u0020e^(2t)\u0020cos(t^2).\u0020Taking\u0020the\u0020derivative\u0020with\u0020respect\u0020to\u0020t:\u000a\u000ady/dt\u0020=\u0020d/dt\u0020(e^(2t)\u0020cos(t^2))\u000a =\u0020e^(2t)\u0020*\u0020d/dt\u0020(cos(t^2))\u0020+\u0020cos(t^2)\u0020*\u0020d/dt\u0020(e^(2t))\u000a\u000aUsing\u0020the\u0020chain\u0020rule,\u0020we\u0020have:\u000a\u000ad/dt\u0020(cos(t^2))\u0020=\u0020-sin(t^2)\u0020*\u0020d/dt\u0020(t^2)\u0020=\u0020-2t\u0020*\u0020sin(t^2)\u000a\u000ad/dt\u0020(e^(2t))\u0020=\u00202e^(2t)\u000a\u000aTherefore,\u000a\u000ady/dt\u0020=\u0020e^(2t)\u0020*\u0020(-2t\u0020*\u0020sin(t^2))\u0020+\u0020cos(t^2)\u0020*\u00202e^(2t)\u000a =\u0020-2te^(2t)sin(t^2)\u0020+\u00202e^(2t)cos(t^2)\u000a =\u00202e^(2t)(cos(t^2)\u0020-\u0020t\u0020sin(t^2))\u000a\u000aFinally,\u0020let's\u0020find\u0020dy/dx\u0020by\u0020dividing\u0020dy/dt\u0020by\u0020dx/dt:\u000a\u000ady/dx\u0020=\u0020(2e^(2t)(cos(t^2)\u0020-\u0020t\u0020sin(t^2)))\u0020/\u0020(4t\u0020cos(2t^2)\u0020+\u00201)