Question

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

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

Hint:

If $z=tan(y+ax)+\sqrty-ax$, then $zxx-azyy$ is equal to

Answer 1

The Correct Option is A

Step 1: Concept
Partial differentiation involves differentiating a function with respect to one variable while holding others constant.
Step 2: Analysis
Differentiating $z$ twice with respect to $x$ ($z_{xx}$) and twice with respect to $y$ ($z_{yy}$).
Step 3: Evaluation
The second-order partial derivatives are calculated:
$z_{xx} = 2a^{2}sec^{2}(y+ax)tan(y+ax) - \frac{a^{2}}{4}(y-ax)^{-3/2}$
$z_{yy} = 2sec^{2}(y+ax)tan(y+ax) - \frac{1}{4}(y-ax)^{-3/2}$
Step 4: Conclusion
Multiplying $z_{yy}$ by $a^{2}$ makes it identical to $z_{xx}$. Therefore, $z_{xx} - a^{2}z_{yy} = 0$.
Final Answer: (a)