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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.

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)

knowledge of the rules of differentiating.

I am going to list two different attempts, one implicit and one explicit.

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.

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.

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)

**Question 1****1. Homework Statement***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. Homework 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]**

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 http://www.coolmath.com/graphit/index.html" [Broken] 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.**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 http://www.coolmath.com/graphit/index.html" [Broken] 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.

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