$f(x,y) = \ln (1-x^2-y^2)$. What's the derivative?

Question

What's is the derivative of $f(x,y)= \ln(1-x^2-y^2)$?

I got it to $f_x(x,y)= 2x^-1$ and the same for $f_y(x,y)$, but that's obviously incorrect. How do I solve this problem properly?

You need to do chain rule to differentiate $\ln$ first. – Jan 08 '16 at 11:20 — , Jan 08 '16 at 11:20
fx(x,y)= 2x/(1-x^2-y^2) Is this correct? – Emil Jan 08 '16 at 11:27 — Emil, Jan 08 '16 at 11:27

score 1 · Answer 1 · answered Jan 08 '16 at 11:24

1

You have to use the chain rule,

$$f_x(x,y) = \frac{-2x}{1-x^2-y^2}$$

$$f_y(x,y) = \frac{-2y}{1-x^2-y^2}$$

answered Jan 08 '16 at 11:24

fosho

1 Answers1