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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luckyb0y1314's profile pic

Posted on (Answer #1)


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.

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^2

Now, 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.

sciencesolve's profile pic

Posted on (Answer #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).`

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