Differentiation Question 98
Question: If $ z=\sec (y-ax)+\tan (y+ax), $ then $ \frac{{{\partial }^{2}}z}{\partial x^{2}}-a^{2}\frac{{{\partial }^{2}}z}{\partial y^{2}}= $
[EAMCET 2002]
Options:
A) z
B) 2z
C) 0
D) -z
Show Answer
Answer:
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) $
$ +2a^{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 $
BETA
