a) i) find dy/dx of y = 3x^4 - 8x^3 - 3 ii) then find d^2y/dx^2 b) verify that x=2 at a stationary point on the curve c c) is this point a minima or a maxima

So the first step is to find the first differential of yTo do this we simply have to differentiate:so with C being our coefficient, n being the power that x is.we need to reduce the power of x by 1, whilst multiplying our coefficient C by n. So what does this look like? We'll start with the first part.we have 3x^4 which is cx^n, so to differentiate we must reduce the power and multiple the coefficient. Giving ncx^(n-1), in terms of our question this is (3*4)x^(4-1) giving 12x^3.now if we do the same for the rest we get dy/dx = 12x^3 - 24x^2, but what about the 3?The three in this case has no x in it, as such we can't reduce the power. Instead we simply reduce this factor to 0.In the second part we are asked to find the second differential of y. To do this we simply have to differentiate dy/dx this time.So repeating the process above again, we would get12x^3 - 24x^2 becoming 36x^2 - 48xNow for part b, we need to show that when x = 2 this is a stationary point of our curve.so we've just being playing with our y and differentiating it, the first time we did this we got the rate of change of y with respect to x, or as this is more commonly known we got the gradient. Now if a point is stationary its no longer moving, so its rate of change is equal to 0. So for us, if our gradient is equal to zero then we will find our stationary point.so earlier we found dy/dx to be equal to 12x^3 - 24x^2if we take x = 2 and place this into our equal we can check to test if at x = 2 we have a stationary point.12(2)^3 - 24(2)^2 = 12(8) - 24(4) = 96 - 96 = 0So we have verified that at x = 2, it is indeed a stationary point on our curve.For the final part of the question we have been asked to determine if this stationary point is a minima or a maximaTo do this its very simple and much like the last part of the question, this time instead of putting x = 2 into the first differential the gradient, we put it into our second differential, the rate of change of the rate of change.so this is x = 2 into 36x^2 - 48x, 36(2)^2 - 48(2) = 36(4) - 48(2) = 144 - 96 = 48Now if this value is positive, we have a minima and if it is negative we have a maxima. So in our case as this value is positive we have a minima stationary point.