# Homework Help: Higher Derivative Question

1. Oct 22, 2006

### 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.

2. Oct 22, 2006

### neutrino

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

3. Oct 22, 2006

### Office_Shredder

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

4. Oct 22, 2006

### 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.

Last edited: Oct 22, 2006
5. Oct 22, 2006

### Office_Shredder

Staff Emeritus
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'

6. Oct 22, 2006

### neutrino

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.

7. Oct 22, 2006

### 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

8. Oct 22, 2006

### 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.

Last edited: Oct 22, 2006
9. Oct 22, 2006

### 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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook