PDA

View Full Version : Differentiating y"


Gernald
Feb 19, 2009, 09:54 AM
So I'm a little confused about what to do with a in this equation and am not sure if it should be 0 or 4a^3

I'm trying to find y" of x^4+y^4=a^4 and so far I have 4x^3+4y^3 x y' =?? 4a^3 or zero.

Once I know what the a^4 is supposed to be I can solve it the rest of the way but I have no idea what to do with it.

Thanks

ebaines
Feb 19, 2009, 10:14 AM
a is a constant, right? What's the derivative of a constant? Asked another way - do constants change with a change in x?

You should check your differentiation - the first derivative of y^4 would be 4 y^3 y'. Not sure why you have an extra x in there.

Gernald
Feb 19, 2009, 10:32 AM
a is a constant, right? What's the derivative of a constant? Asked another way - do constants change with a change in x?

You should check your differentiation - the first derivative of y^4 would be 4 y^3 y'. Not sure why you have an extra x in there.

the x was in there to show multiplication.
The dirivitive of a constant is zero, but I wasn't sure if it's still zero if the constant is taken to a power.

ebaines
Feb 19, 2009, 10:57 AM
The dirivitive of a constant is zero, but I wasn't sure if it's still zero if the constant is taken to a power.


A constant raised to a constant power is a contstant.

Gernald
Feb 19, 2009, 11:53 AM
huh?? So it's ^4??

ebaines
Feb 19, 2009, 12:13 PM
What I was saying is that the derivative of a constant is 0. a^4 is a constant, so its derivative is... 0.