Question

given the equation: x^4+y^4=16. find dy/dx

Answer 1

First you need to solve in order to Y.

y^4 = 16-x^4
y= sqrt(sqrt(16-x^4))
now you just need to derivate:

dy/dx = -x^3/(16-x^4)^(3/4)

Answer 2

what is the d^2y/dx^2

Answer 3

-(48 x^2)/(16-x^4)^(7/4)

Answer 4

how did you get that?

Answer 5

try this
\[4x^3+4y^3y'=0\]
\[y'=-\frac{x^3}{y^3}\]

Answer 6

don't need to solve for y if you do not mind having a y in your answer

Answer 7

That is correct.

Answer 8

dy/dx(x^4)+dy/dx(y^4)=dy/dx(16)

4x^3+4y^3.dy/dx=0
dy/dx=-4x^3/4y^3
dy/dx=-x^3/y^3