Question

If $ u=\log ({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz), $ then $ (x,y,z)\left( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} \right) $ is equal to:

• A. 0
• B. 1
• C. $ u $
• D. 3
• E. $ -1 $

Answer 1

Answer: (D)

$ \because $ $ u=\log ({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz) $ On differentiating w. r. t. $ x,y,z $ respectively, we get $ \frac{\partial u}{\partial x}=\frac{1}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}(3{{x}^{2}}-3yz) $ $ \frac{\partial u}{\partial y}=\frac{1}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}(3{{y}^{2}}-3xz) $ and $ \frac{\partial u}{\partial z}=\frac{1}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}(3{{z}^{2}}-3xy) $ $ \therefore $ $ (x+y+z)\left( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} \right) $ $ =\frac{3(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}{({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)} $ $ =\frac{3({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)}{({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)}=3 $