STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

$\frac{{{\partial ^2}z}}{{\partial {x^2}}} = {a^2}{\sec ^3}(y - ax) + {a^2}\sec (y - ax){\tan ^2}(y - ax)$

$ + 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 $ $\frac{{{\partial ^2}z}}{{\partial {x^2}}} - {a^2}\frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0$