Question

If $(x+y)\sin u=x^2{y}^2$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$ is equal to:

• A. $\sin u$
• B. $\mathrm{cosec}u$
• C. $\tan u$
• D. $3\tan u$

Answer 1

The correct option is D

$3\tan u$

Explanation for the correct option.

Step 1: Simplify the given equation.

$\begin{array}{rcl}(x+y)\sin u& =& x^2{y}^2\\ \Rightarrow \sin u& =& \frac{x^2{y}^2}{x+y}\\ & =& \frac{x^4\left({\displaystyle \frac{y^2}{x^2}}\right)}{x\left(1+{\displaystyle \frac{y}{x}}\right)}\\ & =& \frac{x^3\left(\frac{y^2}{x^2}\right)}{1+\frac{y}{x}}\\ & =& x^{3}f\left(\frac{y}{x}\right)\end{array}$

Now, this is homogenous function with $n=3$.

Step 2: Apply Euler's Theorem.

$x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}=nv$ [By Euler's Theorem]

Let $v=\sin u$

$\begin{array}{rcl}\Rightarrow x\frac{\partial \left(\sin u\right)}{\partial x}+y\frac{\partial \left(\sin u\right)}{\partial y}& =& 3\left(\sin u\right)\\ & \Rightarrow & x\cos u\frac{\partial u}{\partial x}+y\cos u\frac{\partial u}{\partial x}=3\left(\sin u\right)\\ \Rightarrow \cos u\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}\right)& =& 3\left(\sin u\right)\\ \Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}& =& 3\tan u\end{array}$

Hence, option D is correct.

Alternatively,

Step 1: Differentiate partially with respect to $x$

The given equation is $(x+y)\sin u=x^2{y}^2$. On differentiating partially with respect to $x$, we get

$\left(1+0\right)\sin u+\left(x+y\right)\cos u\frac{\partial u}{\partial x}=2x{y}^2$

By multiplying both sides by $x$, we get

$x\sin u+x\left(x+xy\right)\cos u\frac{\partial u}{\partial x}=2x^2{y}^2.......\left(1\right)$

Step 2: Differentiate partially with respect to $y$

The given equation is $(x+y)\sin u=x^2{y}^2$. On differentiating partially with respect to $y$, we get

$\left(0+1\right)\sin u+\left(x+y\right)\cos u\frac{\partial u}{\partial x}=2x^{2}y$

By multiplying both sides by $y$, we get

$y\sin u+y\left(x+y\right)\cos u\frac{\partial u}{\partial x}=2x^2{y}^2.......\left(2\right)$

Step 3: Find the value of $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$

On adding $\left(1\right)\mathrm{and}\left(2\right)$, we get

$\begin{array}{rcl}\left(x+y\right)\sin u+\left(x+y\right)\cos u\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}\right)& =& 4x^2{y}^2\\ \Rightarrow x^2{y}^2+\left(x+y\right)\cos u\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}\right)& =& 4x^2{y}^2\left[\mathrm{Given}(x+y)\sin u=x^2{y}^2\right]\\ \Rightarrow \left(x+y\right)\cos u\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}\right)& =& 3x^2{y}^2\\ \Rightarrow \left(x+y\right)\cos u\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}\right)& =& 3(x+y)\sin u\\ \Rightarrow \cos u\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}\right)& =& 3\sin u\\ \Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial x}& =& 3\tan u\end{array}$

Hence, option D is correct.