How do I find the second derivative of a higher order function?

[loadsy]

Alright I decided I'd create a new topic just because the other one was getting fairly lengthy.

I'm having trouble with the following "Higher Derivative" question

It states, find y'' by implicit differentiation.

x^4 + y^4 = a^4

d/dx (x^4+y^4) = d/dx (a^4)
4x^3 + 4y^3 dy/dx = 4a^3

However, what is the next step from here, I thought perhaps cancelling out all the 4's and leaving it as:
x^3 + y^3 dy/dx = a^3
and bring the x^3 over so it's:
a^3-x^3 = y^3 dy/dx
dy/dx = a^3-x^3 / y^3

However, I'm not positive on that because I was told from a friend that 4a^3 is a constant so it can be equal to 0, thus the dy/dx would be -x^3/y^3.

Am I on the right track here or is there an easier way of solving this equation. Thanks a lot guys.

 

[neutrino]

If 'a' is a constant, then its derivative (with respect to any variable) is zero.

 

[Office_Shredder]

However, I'm not positive on that because I was told from a friend that 4a^3 is a constant so it can be equal to 0, thus the dy/dx would be -x^3/y^3.

4a^3 can't just be written as 0. But as neutrino pointed out, the derivative of a^4 isn't 4a^3

 

[loadsy]

Ahhh I think I'm following you, so pretty much from the first step you could automatically assume that a^4 it is equal to 0 when you take the derivative, and disregard the 4a^3 because that's not possible since it is a constant, thus after solving the problem it would be dy/dx = -x^3/y^3, but that's for y', not y''. You need to take the derivative of that now if I'm not mistaken using the rule:

f'g-fg' / g^2

And sub in:

-x^3(y^3) - y^3(-x^3)
---------------------------
(y^3)^2

And take the derivatives.

 

[Office_Shredder]

Correct. In fact, I'm sure you'll notice that once you use the chain rule, you'll get y'' in terms of x, y, and y' All you need to do is substitute -x³/y³ for y'

 

[neutrino]

Ahhh I think I'm following you, so pretty much from the first step you could automatically assume that a^4 it is equal to 0 when you take the derivative, and disregard the 4a^3 because that's not possible since it is a constant, thus after solving the problem it would be dy/dx = -x^3/y^3, but that's for y', not y''. You need to take the derivative of that now if I'm not mistaken.

You're right, and you can also take the derivative x^3 + y^3 dy/dx = 0, using the product rule for the second term.

 

[HallsofIvy]

Notice that when you differentiated y^4 with respect to x you did not get "4y^3", you got "4y^3 y' ". You could say that the derivative of a^4 with respect to x is 4a^3 a' and then since a is a constant, a'= 0 so (a^4)'= 4a^3(0)= 0. Of course, it's simpler to argue that a^4 is itself a constant so (a^4)'= 0 immediately.

As neutrino said, once you have x^2+ y^3 y'= 0 you can just continue using "implicit differentiation": 2x+ 3y^2 (y')^2+ y^3 y"= 0

 

[loadsy]

Oh wow, alright thanks a lot for the advice all of you. HallsofIvy clarified that constant business for me perfectly. I understand that a whole lot better now. Except one correction in your description, it's x^3+y^3 y' = 0 believe as neutrino suggested, not x^2+ y^3 y'= 0
Hence, it would be:
3x^2 + 3y^2 (y')^2 + y^3 y'' = 0

And then for y' you'd sub in -x^3/y^3 as Office Shredder suggested, and solve for y''?

So it would be 3x^2 + 3y^2(-x^3/y^3)^2 + y^3 y'' = 0 I think anyways.

 

[loadsy]

Alright here's an update on what I did for this question anyways, because I assume you use the rule: f'g-fg'/g^2
so from -x^3/y^3 to differentiate this expression I went:

-3x^2(y^3) - (-x^3)(3y^2)y' / (y^3)^2
= -3x^2(y^3) - (-x^3)(3y^2)(-x^3/y^3) / (y^3)^2
Now how do I factor this expression.