Example of product rule - if y=e^(3x-x^3), what are the coordinates of stationary points and what are their nature?
Given this is a maths problem a whiteboard will be heavily relied upon, but I'll do my best to get the appropriate notation here.Starting with y=e3x-x^3, we have to get to the stationary points and find whether they're a maximum or minimum. This involves differentiating twice, which will in this instance necessitate the use of the chain rule and then the product rule. In this example, let's let 3x-x3=u, netting us y=eu , du/dx=3-3x2, and dy/du=eu still due to the nature of differentiating e. Multiplying dy/du and du/dx nets us dy/dx, in this case dy/dx=(e3x-x^3)(3-3x2). To find the stationary points, we make this equal to zero, and we can see the valid points are x=1 and x=-1. The coordinates of the stationary points are, subbing back into the original y, (1, e2) and (-1, e-2).Now that we have these values, we can find their nature as maxima or minima by differentiating dy/dx to d2y/dx2. This involves the product rule, multiplying a bracket from the dy/dx earlier with the differential of the other bracket, and adding it to the other way around. ie, if y=vu (v and u being expressions with x), dy/dx=v(du/dx)+u(dv/dx). In this case (once again I'd ordinarily be going through this in detailed steps on the whiteboard) this nets us d2y/dx2= (e3x-x^3)(-6x)+(e3x-x^3)(3-3x2)2. Subbing in our earlier x values of 1 and -1 gives us d2y/dx2 values of -6e2, a maximum point, and 6e-2, a minimum point, respectively.
