# Find the second derivative using implicit differentiation. `x^3-y^3=3`Find x^3-y^3=3, use Implicit Differentiation to find the second derivative (d^2 y/dx^2). Show all work please.

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x^3-y^3=3

Here's a point we need to think about.**If y= x^n, dy/dx= nx^(n-1), where n is real number.**x^3 looks similar with the x^n above. Isn't it? Then, we can use the formula to carry out differentiation.

How about y^3? To solve this, we need to know the concept of chain rule.

Consider dy/dx= (dy/du)*(du/dx),

by letting u= y^3.

du/dx= (dy/dx)/(dy/du)

du/dx= (dy/dx)*(du/dy)

d(y^3)/dx = (dy/dx)* [d(y^3)dy]

d(y^3)/dx = (3y^2)*(dy/dx)

Using implicit differentiation to find the first derivative.

3x^2-(3y^2)(dy/dx)=0

Bring over the 3y^2(dy/dx) to right hand side, so we have

3x^2= 3y^2 (dy/dx)

Cancel the 3 both side, therefore we get

x^2= y^2 (dy/dx).

Notice that y^2 (dy/dx) is a product. So, we use the product rule in solving the RHS by letting u= y^2 and v= dy/dx.

The formula for product rule is **dy/dx = u(dv/dx)+v(du/dx).**

Let u=y^2

du/dx=2y (dy/dx)

Let v=dy/dx

dv/dx= d^2 y/dx^2Now, we do the differentiation one more time to find the second derivative. For the part in left hand side, u need to perform differentiation using the formula dy/dx = nx^(n-1) since the x^2 is an example of y= x^n. For the part in right hand side, u just substitute in the elements according to the formula for product rule.

2x = y^2 (d^2 y/dx^2) + (dy/dx) [ 2y(dy/dx)]

Rewrite it:

we have

y^2 (d^2 y/ dx^2)= 2x- 2y(dy/dx)^2

d^2 y/ dx^2= [2x-2y (dy/dx)^2] / y^2

d^2 y/ dx^2= 2[x-y (dy/dx)^2] / y^2

Therefore, the second derivative for x^3-y^3=3 is 2 [x-y (dy/dx)^2] / y^2.

You need to write the function in terms of x, hence, you need to isolate `y^3` to the left side such that:

`-y^3 = 3 - x^3 => y^3 = x^3 - 3 => y = root(3)(x^3 - 3)`

You need to convert the cube root into a power such that:

`y = (x^3 - 3)^(1/3)`

You need to differentiate the function with respect to x, using the chain rule, such that:

`(dy)/(dx) = (1/3)(x^3 - 3)^(1/3 - 1)(3x^2)`

`(dy)/(dx) = (1/3)(x^3 - 3)^(-2/3)(3x^2)`

`(dy)/(dx) = (x^2)(x^3 - 3)^(-2/3)`

You need to differentiate again to evaluate `(d^2y)/(dx^2)` such that:

`(d^2y)/(dx^2)= (x^2)'((x^3 - 3)^(-2/3)) + (x^2)((x^3 - 3)^(-2/3))'`

`(d^2y)/(dx^2) = 2x((x^3 - 3)^(-2/3)) + (x^2)((x^3 - 3)^(-2/3 - 1))*(3x^2)`

`(d^2y)/(dx^2) = 2x((x^3 - 3)^(-2/3)) + 3x^4((x^3 - 3)^(-5/3))`

You should factor out `x((x^3 - 3)^(-5/3))` such that:

`(d^2y)/(dx^2) = (x((x^3 - 3)^(-5/3)))(2((x^3 - 3)^(-2/3+5/3)) + 3x^3)`

`(d^2y)/(dx^2) = (x((x^3 - 3)^(-5/3)))(2(x^3 - 3 + 3x^3))`

`(d^2y)/(dx^2) = ((x^3 - 3)^(-5/3)))(8x^4 - 6x)`

**Hence, evaluating the second derivative using implicit differentiation yields `(d^2y)/(dx^2) = ((x^3 - 3)^(-5/3)))(8x^4 - 6x).` **