If \(z=ln(\frac{x^2+y^2}{x+y})-e^{\frac{x^2+y^2}{x+y}}\) then find \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}\).
Posted: Thu Jul 14, 2022 9:29 am
a) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
b) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
c) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1+\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
d) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
b) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
c) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1+\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
d) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)