xt=12 sin2t^2+t yt=e^2t cost^2 find the dydx

To find dy/dx, we need to differentiate y(t) with respect to x(t).

First, let's find dx/dt. We have x(t) = 1/2 sin(2t^2) + t. Taking the derivative with respect to t:

dx/dt = d/dt (1/2 sin(2t^2) + t) = d/dt (1/2 sin(2t^2)) + d/dt (t) = cos(2t^2) * d/dt (2t^2) + 1 = cos(2t^2) * 4t + 1 = 4t cos(2t^2) + 1

Now, let's find dy/dt. We have y(t) = e^(2t) cos(t^2). Taking the derivative with respect to t:

dy/dt = d/dt (e^(2t) cos(t^2)) = e^(2t) * d/dt (cos(t^2)) + cos(t^2) * d/dt (e^(2t))

Using the chain rule, we have:

d/dt (cos(t^2)) = -sin(t^2) * d/dt (t^2) = -2t * sin(t^2)

d/dt (e^(2t)) = 2e^(2t)

Therefore,

dy/dt = e^(2t) * (-2t * sin(t^2)) + cos(t^2) * 2e^(2t) = -2te^(2t)sin(t^2) + 2e^(2t)cos(t^2) = 2e^(2t)(cos(t^2) - t sin(t^2))

Finally, let's find dy/dx by dividing dy/dt by dx/dt:

dy/dx = (2e^(2t)(cos(t^2) - t sin(t^2))) / (4t cos(2t^2) + 1