Question

If x^2+y^2=16 y^1 equals ?

Answer 1

derivative, not solving for y

Answer 2

You need to do implicit differentiation:
\[2x+2yy'=16y' \implies y'=\frac{2x}{16-2y}=\frac{x}{8-y}.\]

Answer 3

But the derivative of a constant is 0

Answer 4

So the result should be -x/y

Answer 5

@mr.math x^2+y^2=16 , derivative of y equals?

Answer 6

Oh I thought it was \(x^2+y^2=16y\). If the problem is as you stated in your last comment, then:
\(2x+2yy'=0 \implies y'=-\frac{x}{y}\), as Mertsj said.

Answer 7

Thank you