Q99 / 200 · TS EAMCET 2002 · Mathematics
If $z=\frac{y}{x}\left[\sin \frac{x}{y}+\cos \left(1+\frac{y}{x}\right)\right]$, then $x \frac{\partial z}{\partial x}$ is equal to
- A$y \frac{\partial z}{\partial y}$
- Correct answerB$-y \frac{\partial z}{\partial y}$
- C$2 y \frac{\partial z}{\partial y}$
- D$2 y \frac{\partial z}{\partial x}$
Solution
We have,
$$
z=\frac{y}{x}\left[\sin \frac{x}{y}+\cos \left(1+\frac{y}{x}\right)\right]
$$
$$
\begin{array}{r}
\frac{\partial z}{\partial x}=\frac{y}{x}\left[\cos \frac{x}{y} \cdot \frac{1}{y}-\sin \left(1+\frac{y}{x}\right)\left(-\frac{y}{x^2}\right)\right] \\
+\left(-\frac{y}{x^2}\right)\left[\sin \frac{x}{y}+\cos \left(1+\frac{y}{x}\right)\right] \\
x \frac{\partial z}{\partial x}=y\left[\frac{1}{y} \cos \frac{x}{y}+\frac{y}{x^2} \sin \left(1+\frac{y}{x}\right)\right] \\
-\frac{y}{x}\left[\sin \frac{x}{y}+\cos \left(1+\frac{y}{x}\right)\right] \\
\Rightarrow \quad x \frac{\partial z}{\partial x}=\cos \frac{x}{y}+\frac{y^2}{x^2} \sin \left(1+\frac{y}{x}\right)-z
\end{array}
$$
Similarly,
$$
y \frac{\partial z}{\partial y}=-\cos \frac{x}{y}-\frac{y^2}{x^2} \sin \left(1+\frac{y}{x}\right)+z
$$
On adding Eqs. (i) and (ii)
$$
x \frac{\partial z}{\partial x}=-y \frac{\partial z}{\partial y}
$$
Hence, the correct answer is option: 'B'.