z=tan (y+a x)+sqrt{y-a x} Rightarrow z_{x x}- | vedclass

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(A) Given, $z=	an (y+a x)+\sqrt{y-a x}$ First, find the partial derivatives with respect to $x$: $z_x = \sec^2(y+ax) \cdot a + \frac{1}{2\sqrt{y-ax}}

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(A) Given, $z=	an (y+a x)+\sqrt{y-a x}$ First, find the partial derivatives with respect to $x$: $z_x = \sec^2(y+ax) \cdot a + \frac{1}{2\sqrt{y-ax}} \cdot (-a)$ $z_{xx} = \frac{\partial}{\partial x} [a \sec^2(y+ax) - \frac{a}{2\sqrt{y-ax}}] = 2a^2 \sec^2(y+ax) 	an(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}$ Next, find the partial derivatives with respect to $y$: $z_y = \sec^2(y+ax) + \frac{1}{2\sqrt{y-ax}}$ $z_{yy} = \frac{\partial}{\partial y} [\sec^2(y+ax) + \frac{1}{2\sqrt{y-ax}}] = 2 \sec^2(y+ax) 	an(y+ax) - \frac{1}{4(y-ax)^{3/2}}$ Now, calculate $z_{xx} - a^2 z_{yy}$: $z_{xx} - a^2 z_{yy} = [2a^2 \sec^2(y+ax) 	an(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}] - a^2 [2 \sec^2(y+ax) 	an(y+ax) - \frac{1}{4(y-ax)^{3/2}}]$ $z_{xx} - a^2 z_{yy} = 2a^2 \sec^2(y+ax) 	an(y+ax) - \frac{a^2}{4(y-ax)^{3/2}} - 2a^2 \sec^2(y+ax) 	an(y+ax) + \frac{a^2}{4(y-ax)^{3/2}} = 0$

$z=\tan (y+a x)+\sqrt{y-a x} \Rightarrow z_{x x}-a^2 z_{y y}$ is equal to

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