Question

If $z=\tan \left(y+ax\right)-\sqrt{y-ax}$, then $z_{xx}-a^2{z}_{yy}$ is equal to

• A. $0$
• B. $2$
• C. $z_x+z_y$
• D. $z_x{z}_y$

Answer 1

The correct option is A

$0$

Explanation for the correct expression.

Step 1. Find the value of $z_{xx}$.

Partially differentiate $z=\tan \left(y+ax\right)-\sqrt{y-ax}$ with respect to $x$.

$\begin{array}{rcl}{z}_x& =& \frac{\partial }{\partial x}\left(\tan \left(y+ax\right)-\sqrt{y-ax}\right)\\ & =& {\sec }^2\left(y+ax\right)\times \left(0+a\right)-\frac{1}{2\sqrt{y-ax}}(0-a)\\ & =& a{\sec }^2\left(y+ax\right)+\frac{a}{2\sqrt{y-ax}}\end{array}$

Now partially differentiate $z_x\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & a\end{array}\begin{array}{rcl}& & \sec \end{array}\begin{array}{rcl}& & \end{array}$ with respect to $x$.

$\begin{array}{rcl}{z}_{xx}& =& \frac{\partial }{\partial x}\left(a{\sec }^2\left(y+ax\right)+\frac{a}{2\sqrt{y-ax}}\right)\\ & =& a2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)\times \left(0+a\right)+\frac{a}{2}\frac{-1}{2}{\left(y-ax\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(0-a)\\ & =& 2a^2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)-\frac{a^2}{4}{\left(y-ax\right)}^{3/2}\end{array}$

Step 2. Find the value of $z_{yy}$.

Partially differentiate $z=\tan \left(y+ax\right)-\sqrt{y-ax}$ with respect to $y$.

$\begin{array}{rcl}{z}_y& =& \frac{\partial }{\partial y}\left(\tan \left(y+ax\right)-\sqrt{y-ax}\right)\\ & =& {\sec }^2\left(y+ax\right)\times \left(1+0\right)-\frac{1}{2\sqrt{y-ax}}(1+0)\\ & =& {\sec }^2\left(y+ax\right)-\frac{1}{2\sqrt{y-ax}}\end{array}$

Now partially differentiate $z_y\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \sec \end{array}\begin{array}{rcl}& & \end{array}$ with respect to $y$.

$\begin{array}{rcl}{z}_{yy}& =& \frac{\partial }{\partial y}\left({\sec }^2\left(y+ax\right)+\frac{1}{2\sqrt{y-ax}}\right)\\ & =& 2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)\times \left(1+0\right)+\frac{1}{2}\frac{-1}{2}{\left(y-ax\right)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}(1-0)\\ & =& 2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)-\frac{1}{4}{\left(y-ax\right)}^{3/2}\end{array}$

Step 3. Find the value of the expression.

The value of the expression $z_{xx}-a^2{z}_{yy}$ can be found by substituting the values found.

$\begin{array}{rcl}{z}_{xx}-a^2{z}_{yy}& =& 2a^2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)-\frac{a^2}{4}{\left(y-ax\right)}^{3/2}-a^2\left(2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)-\frac{1}{4}{\left(y-ax\right)}^{3/2}\right)\\ & =& 2a^2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)-\frac{a^2}{4}{\left(y-ax\right)}^{3/2}-2a^2{\sec }^2\left(y+ax\right)\tan \left(y+ax\right)+\frac{a^2}{4}{\left(y-ax\right)}^{3/2}\\ & =& 0\end{array}$

Hence, the correct option is A.