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) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
If \(z=ln(\frac{x^2+y^2}{x+y})-e^{\frac{x^2+y^2}{x+y}}\) then find \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}\).
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answerhappygod
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If \(z=ln(\frac{x^2+y^2}{x+y})-e^{\frac{x^2+y^2}{x+y}}\) then find \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}\).
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