Question

How do you find `(d^2y)/(dx^2)` given `x^2+y^2=13`?

Answer 1

See below.

Explanation:

Given `x^2+y^2=13`,

`2x+2y dy/dx = 0`, so

`dy/dx = (-x)/y`.

Therefore,

`d/dx(dy/dx) = ((-1)y-(-x)dy/dx)/y^2`

`=( -y+x(-x/y))/y^2 * y/y`

`= (-y^2-x^2)/y^3`

`= (-13)/y^3`.