Question

\( z = \tan(y + ax) + \sqrt{y - ax} \implies z_{xx} - a^{2} z_{yy} \) is equal to

• A. 0
• B. 2
• C. $z_{x} + z_{y}$
• D. $z_{x}z_{y}$

Hint:

For any function $f(y \pm ax)$, the second partial derivative $z_{xx}$ is always $a^2 z_{yy}$.

Answer 1

The Correct Option is A

Step 1: Partial Derivatives with respect to x
$z_{x} = a \sec^{2}(y+ax) - \frac{a}{2\sqrt{y-ax}}$. $z_{xx} = 2a^{2} \sec^{2}(y+ax) \tan(y+ax) - \frac{a^{2}}{4(y-ax)^{3/2}}$.
Step 2: Partial Derivatives with respect to y
$z_{y} = \sec^{2}(y+ax) + \frac{1}{2\sqrt{y-ax}}$. $z_{yy} = 2 \sec^{2}(y+ax) \tan(y+ax) - \frac{1}{4(y-ax)^{3/2}}$.
Step 3: Verification
By inspection, $z_{xx} = a^{2} z_{yy}$. Therefore, $z_{xx} - a^{2}z_{yy} = 0$.
Final Answer: (a)