Differentiation Question 75
Question: If $ u=\log (x^{2}+y^{2}), $ then $ \frac{{{\partial }^{2}}u}{\partial x^{2}}+\frac{{{\partial }^{2}}u}{\partial y^{2}}= $
[EAMCET 1994]
Options:
A) $ \frac{1}{x^{2}+y^{2}} $
B) 0
C) $ \frac{x^{2}-y^{2}}{{{(x^{2}+y^{2})}^{2}}} $
D) $ \frac{y^{2}-x^{2}}{{{(x^{2}+y^{2})}^{2}}} $
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Answer:
Correct Answer: B
Solution:
$ u=\log (x^{2}+y^{2})\therefore \frac{\partial u}{\partial x}=\frac{1}{x^{2}+y^{2}}.2x $
$ \frac{{{\partial }^{2}}u}{\partial x^{2}}=\frac{(x^{2}+y^{2}).2-2x.2x}{{{(x^{2}+y^{2})}^{2}}} $
$ =\frac{2(y^{2}-x^{2})}{{{(x^{2}+y^{2})}^{2}}} $ - $ \frac{\partial u}{\partial y}=\frac{1}{x^{2}+y^{2}}.2y $
$ \therefore \frac{{{\partial }^{2}}u}{\partial y^{2}}=\frac{(x^{2}+y^{2}).2-2y.2y}{{{(x^{2}+y^{2})}^{2}}}=\frac{2(x^{2}-y^{2})}{{{(x^{2}+y^{2})}^{2}}} $
$ \therefore $ $ \frac{{{\partial }^{2}}u}{\partial x^{2}}+\frac{{{\partial }^{2}}u}{\partial y^{2}}=0 $ .
BETA
