Question

Find y"in terms of x and y by implicit differentiation.
(x^3)+(y^3)=4^3.
This problem is really stumping me.
So, now i have it down to 3x+3y(-x^2/y^2)

Answer 1

Where is this coming from?
"(-x^2/y^2)"

Answer 2

When you solve for dy/dx

Answer 3

Let's take this term-by-term.
What is d/dx of x^3?

Answer 4

3x^2

Answer 5

What is d/dx of y^3?

Answer 6

3y^2(dy/dx)

Answer 7

Ok, great, and 4^3 is a constant, so the derivative of that is 0.
So you should have so far.
\[\large 3x^2+3y^2y'=0\]

Answer 8

Solve that for y'.

Answer 9

y'=-x^2/y^2

Answer 10

Now you want second derivative, right?
So take d/dx implicitly of that.

Answer 11

You can use quotient rule, or express it as
\[\large -x^2y^{-2}\] and use product rule

Answer 12

So, then it would be -x^2(-2y^-3)+y^-2(-2x) correct?

Answer 13

Once you have that where would you go next?

Answer 14

That looks good, except you're missing the y' in ((-2y'y^-3).
Simplify and substitute in the y' that you already found.

Answer 15

okay so then -x^2(-2y^-3(-x/y))+(y^-2)(2x) right?

Answer 16

y'=-x^2/y^2.

Answer 17

Okay so then -x^2(-2y^-3(-x^2/y^2))+(y^-2)(2x)? Do you go further than this?

Answer 18

You can simplify it down, but that is essentially it.

Answer 19

Okay. Thank you for your help.