TS EAMCET · Maths · Differentiation
\(z=\tan (y+a x)+\sqrt{y-a x} \Rightarrow z_{x x}-a^2 z_{y y} \quad\) is equal to
- A 0
- B 1
- C \(z_x+z_y\)
- D \(z_x z_y\)
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Given, } z=\tan (y+a x)+\sqrt{y-a x} \\ & \Rightarrow \quad z_x=\sec ^2(y+a x) a+\frac{1}{2 \sqrt{y-a x}}(-a) \\ & \Rightarrow \quad z_{x x}=2 \sec ^2(y+a x) \tan (y+a x) a^2 \\ & +\frac{1\left(-a^2\right)}{4(y-a x)^{3 / 2}} \\ & \text { and } \quad…
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