A) z
B) 2z
C) 0
D) ?z
Correct Answer: C
Solution :
\[\frac{\partial z}{\partial x}=-a\sec (y-ax)\tan (y-ax)+a{{\sec }^{2}}(y+ax)\] \[\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}={{a}^{2}}{{\sec }^{3}}(y-ax)+{{a}^{2}}\sec (y-ax){{\tan }^{2}}(y-ax)\] \[+2{{a}^{2}}{{\sec }^{2}}(y+ax)\tan (y+ax)\] \[\frac{\partial z}{\partial y}=\sec (y-ax)\tan (y-ax)+{{\sec }^{2}}(y+ax)\] \[\frac{{{\partial }^{2}}z}{\partial {{y}^{2}}}={{\sec }^{3}}(y-ax)+\sec (y-ax){{\tan }^{2}}(y-ax)\] \[+2{{\sec }^{2}}(y+ax)\tan (y+ax)\] \[\therefore \] \[\frac{{{\partial }^{2}}z}{\partial {{x}^{2}}}-{{a}^{2}}\frac{{{\partial }^{2}}z}{\partial {{y}^{2}}}=0\]
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