FInd the derivative y'

Question

Let y=ln(x^2+y^2). Determine the derivative y' at the point (-sqrte^8-64,8).

y'(-sqrte^8-64)=

Answer 1

Take the derivative implicitly:

y = ln(x² + y²)

e^y = e^(ln(x² + y²))

e^y = x² + y²

Let y' represent dy/dx:

(e^y)' = (x² + y²)'

e^y•y' = 2x + 2y•y'

e^y•y' - 2y•y' = 2x

y'(e^y - 2y) = 2x

y' = (2x)/(e^y - 2y)

Plug in x = -√(e⁸ - 64) and y = 8:

y' = (-2√(e⁸ - 64))/(e⁸ - 2•8)

y' = (-2√(e⁸ - 64))/(e⁸ - 16)

This is the exact value but it is approx -0.0364