How do you find the gradient of a parametric equation at a certain point?

The  easiest way to see this is with an example. So let us take the curve:

    x = cos(2t) ,  y = sin(t) with a point P where t = pi/6

The gradient at any point is equal to the derivative of y with respect to x at the given point. But in this situation we cannot simply take dy/dx. However under closer inspection and treating dy/dx as a fraction we see that dy/dx = (dy/dt)(dt/dx). However, this dt/dx may also present problems, but it also can be seen as 1/(dx/dt) in treating it as a fraction as well.

    So all we need to calculate is dy/dt and dx/dt:

        dy/dt = cos(t)  ,  dx/dt = -2sin(2t)

Now dt/dx = 1/(dx/dt) = -1/(2sin(2t))

Then dy/dx = (dy/dt)(dt/dx) = -cos(t)/2sin(2t)

So we now have the derivative at all points in the curve. To find the gradient at the point P we simply substitute in the value of t and calculate.

    This gives:

        -cos(pi/6)/2sin(pi/3) = -(3^1/2/2)/2(3^1/2/2) = -1/2

Thus the gradient of this parametric equation at P is -1/2. This method can easily be extended and adapted to all equations of parametric form.