Question

If we have an expression as ${\sin ^2}x + 2\cos y + xy = 0$, then $\dfrac{{dy}}{{dx}} = ?$
$\left( 1 \right)\dfrac{{\left( {y + 2\sin x} \right)}}{{\left( {2\sin y + x} \right)}}$
\[\left( 2 \right)\dfrac{{\left( {y + \sin 2x} \right)}}{{\left( {2\sin y - x} \right)}}\]
$\left( 3 \right)\dfrac{{\left( {y + 2\sin x} \right)}}{{\left( {\sin y + x} \right)}}$
$\left( 4 \right)$ None of these

Answer 1

Hint: In order to solve this question, we will apply the differentiation in the given equation and then, we will substitute the respective differentiation value of each term in the obtained expression. After that, we will simplify the obtained expression by using appropriate mathematical operations.

Complete step-by-step solution:
Since, the given expression is ${\sin ^2}x + 2\cos y + xy = 0$.
Now, we will differentiate the whole expression with respect to $x$.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^2}x + 2\cos y + xy} \right) = \dfrac{d}{{dx}}\left( 0 \right)\]
Here, we will imply the different sign to each term to proceed further.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^2}x} \right) + \dfrac{d}{{dx}}\left( {2\cos y} \right) + \dfrac{d}{{dx}}\left( {xy} \right) = 0\]
After that, we will use the rule of differentiation and substitute $2\sin x\cos x$ for the differentiation of ${\sin ^2}x$, $ - 2\sin y\dfrac{{dy}}{{dx}}$ for the differentiation of $2\cos y$, and $y + x\dfrac{{dy}}{{dx}}$ for the differentiation of $xy$ in the above expression. Then, the obtained expression is:
\[ \Rightarrow 2\sin x\cos x - 2\sin y\dfrac{{dy}}{{dx}} + y + x\dfrac{{dy}}{{dx}} = 0\]
Now, we will add $2\sin y\dfrac{{dy}}{{dx}}$ and subtract $x\dfrac{{dy}}{{dx}}$ in the above step as,
 \[ \Rightarrow 2\sin x\cos x - 2\sin y\dfrac{{dy}}{{dx}} + y + x\dfrac{{dy}}{{dx}} + 2\sin y\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}} = 0 + 2\sin y\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}}\]
Here, we will cancel out the equal like terms to simplify the above expression as:
\[ \Rightarrow 2\sin x\cos x + y = 2\sin y\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}}\]
Then, we will use the formula of $\sin 2x$ and substitute $\sin 2x$ for $2\sin x\cos x$ in the obtained expression.
\[ \Rightarrow \sin 2x + y = 2\sin y\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}}\]
Now, we will take the term $\dfrac{{dy}}{{dx}}$ common from the above expression and will proceed further.
\[ \Rightarrow \sin 2x + y = \dfrac{{dy}}{{dx}}\left( {2\sin y - x} \right)\]
After that, we will divide by $2\sin y - x$ each side of the above expression and simplify it.
\[ \Rightarrow \dfrac{{\sin 2x + y}}{{2\sin y - x}} = \dfrac{{\dfrac{{dy}}{{dx}}\left( {2\sin y - x} \right)}}{{2\sin y - x}}\]
Here, we will cancel out the term $2\sin y - x$ from the left side of the above expression to get the final answer.
\[ \Rightarrow \dfrac{{\sin 2x + y}}{{2\sin y - x}} = \dfrac{{dy}}{{dx}}\]
Hence, Option B is the right answer.

Note: Here, we will mention some important factors that are used in the solution of the question.
\[ \Rightarrow \dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\]
\[ \Rightarrow \dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x\]
\[ \Rightarrow \dfrac{{d\left( {uv} \right)}}{{dx}} = \dfrac{{du}}{{dx}} \cdot v + u \cdot \dfrac{{dv}}{{dx}}\]