I have two questions, one about differentiating, and the other is about finding the inflection point on a function. Any help would be greatly appreciated. Question 1 1. The problem statement, all variables and given/known data If ay^3 = x^4, show that d^2y/dx^2 = 4x^2/9ay^2 now, this question can be done by both implicit or explicit differentiation. I can leave the ay^3 = x^4 as it is and start simplifying right away, or I could isolate y by making the equation y = (x^4a^-1)^(1/3) 2. Relevant equations knowledge of the rules of differentiating. 3. The attempt at a solution I am going to list two different attempts, one implicit and one explicit. Implicit ay^3 = x^4 a3y^2 dy/dx = 4x^3 dy/dx = 4x^3/ a3y^2 d^2y/dx^2 = (12x^2((a3y^2)-6ya(4x^3)(dy/dx)) / (3y^2a)^2 d^2y/dx^2 = (36x^2y^2a - 24yax^3(dy/dx) ) / (3y^2a)^2 *I then subbed the previous found value for dy/dx into dy/dx in this equation. d^2y/dx^2 = (36x^2y^2a - 96yax^6/3y^2a) / (3y^2a)^2 d^2y/dx^2 = (36x^2y^2a - 32x^6/y) / 9y^4a^2 I stopped here because I could find no way to make the a^2 at the bottom become just a, similarily the y^4 could not be reduced down to y^2 like in the question. Explicit y = (x^4a^-1)^(1/3) y' = (1/3)(x^4a^-1)^(-2/3)(4x^3a^-1) y' = [(x^4a^-1)^(-2/3)(4x^3a^-1)]/3 y'' = [3(12x^2a^-1)^(-2/3)(x^4a^-1)^(-5/3)(4x^3a^-1)]/9 y'' = [-2(48x^5(a^-2)(x^4a^-1)^(-5/3)]/9 y'' = [-96xa^-1(x^4a^-1)(x^4a^-1)^(-5/3)]/9 y'' = -96x / 9a [(x^4/a^1)^(1/3)]^2 since (x^4/a^1)^(1/3) = y y'' = -96x / 9ay^2 as you can see, I got a lot further by the final answer is still wrong... I have part of the answer but have a -96x instead of 4x... could someone please be kind enough to solve this question, or tell me where I went wrong... if it is possible please also provide a solution to the implicit way of solving this question. Question 2 1. Find the point of inflection of y = x^4 - 4x^3 + 6x^2 + 12x at a glance, this question may seem very easy, but for some reason I can't get the right answer (-0.4 and 2.4) 2. to find inflection point, you find the 2nd derivative of the function and solve for x/b] 3. The attempt at a solution y = x^4 - 4x^3 + 6x^2 + 12x y' = 4x^3 - 12x^2 + 12x + 12 y'' = 12x^2 - 24x + 12 let y'' = 0 0 = 12(x^2 - 2x + 1) 0 = 12(x-1)(x-1) x = 1 I put this function into a graphing calculator, it showed no apparent change in concavity at x = 1. I set the min y = -100 and max y = 100 while keeping min x = -20 and max x = 20 to get a better view. the graph seemed to be a parabola but like the answer suggested, the concavity is slightly different between -0.4 and 2.4. anyone wishing to see the graph can just use this one and paste in the function. remember the set the min y, max y, min x, max x values so that you can see the graph as described by me. Thanks for you time.