Question

If ${\sin }^{2}x+2\cos y+xy=0$, then $\frac{dy}{dx}$

• A. $\frac{(y+2\sin x)}{(2\sin y+x)}$
• B. $\frac{(y+\sin 2x)}{(2\sin y-x)}$
• C. $\frac{(y+\sin 2x)}{(\sin y+x)}$
• D. none of these

Answer 1

The correct option is B

$\frac{(y+\sin 2x)}{(2\sin y-x)}$

Explanation for the correct option.

Step 1: Differentiate both sides with respect to $x$ we have:

Given that, ${\sin }^{2}x+2\cos y+xy=0$
$\begin{array}{rcl}2\sin x\cos x-2\sin y\frac{dy}{dx}+\left(y+x\frac{dy}{dx}\right)& =& 0\\ \frac{dy}{dx}(x-2\sin y)& =& (-y-2\sin x\cos x)\\ \frac{dy}{dx}(x-2\sin y)& =& (-y-\sin 2x)\end{array}$

Step 2: Calculate the value of $\begin{array}{rcl}& & \frac{dy}{dx}\end{array}$

$\begin{array}{rcl}\therefore \frac{dy}{dx}& =& \frac{(-y-\sin 2x)}{(x-2\sin y)}\\ & =& \frac{-(y+\sin 2x)}{(x-2\sin y)}\\ & =& \frac{(y+\sin 2x)}{(2\sin y-x)}\end{array}$
Hence option B is correct.