Question

If $\sin \left( xy \right)+\dfrac{x}{y}={{x}^{2}}-y$, then $\dfrac{dy}{dx}=$
1. $\dfrac{y\left( 2xy-{{x}^{2}}\cos xy-1 \right)}{\left( x{{y}^{2}}\cos xy-x+{{y}^{2}} \right)}$
2. $\dfrac{\left( 2xy-{{y}^{2}}\cos xy-1 \right)}{\left( x{{y}^{2}}\cos xy-x+{{y}^{2}} \right)}$
3. $\dfrac{y\left( 2xy-{{y}^{2}}\cos xy-1 \right)}{\left( x{{y}^{2}}\cos xy-x+{{y}^{2}} \right)}$
4. none of these

Answer 1

To solve the given question we will use the concept of implicit differentiation. We will differentiate both sides of the given equation with respect to x. In order to differentiate the equation we will use the power rule and product rule of the differentiation.

Complete step by step answer:
We have been given an equation $\sin \left( xy \right)+\dfrac{x}{y}={{x}^{2}}-y$.
We have to differentiate the given equation.
Now, differentiating the given equation with respect to x we will get
$\Rightarrow \dfrac{d}{dx}\sin \left( xy \right)+\dfrac{d}{dx}\dfrac{x}{y}=\dfrac{d}{dx}{{x}^{2}}-\dfrac{d}{dx}y$
Now, we know that
$\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Now, applying the differentiation rules to the above obtained equation we will get
$\Rightarrow \cos \left( xy \right)\dfrac{d}{dx}\left( xy \right)+\dfrac{d}{dx}\dfrac{x}{y}=2{{x}^{2-1}}-\dfrac{dy}{dx}$
Now, we know that $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$
Now, applying the differentiation rules to the above obtained equation we will get
$\Rightarrow \cos \left( xy \right)\left[ x\dfrac{dy}{dx}+y \right]+\dfrac{dy}{dx}\dfrac{1}{x}-\dfrac{y}{{{x}^{2}}}=2x-\dfrac{dy}{dx}$
Now, simplifying the above obtained equation we will get
$\Rightarrow x\cos \left( xy \right)\dfrac{dy}{dx}+y\cos \left( xy \right)+\dfrac{dy}{dx}\dfrac{1}{x}-\dfrac{y}{{{x}^{2}}}=2x-\dfrac{dy}{dx}$
Now, to find the value of $\dfrac{dy}{dx}$ we need to rearrange the terms in the above obtained equation. Then we will get

& \Rightarrow x\cos \left( xy \right)\dfrac{dy}{dx}+\dfrac{dy}{dx}+\dfrac{dy}{dx}\dfrac{1}{x}=2x-y\cos \left( xy \right)+\dfrac{y}{{{x}^{2}}} \\\ & \Rightarrow \dfrac{dy}{dx}\left[ \cos \left( xy \right)+1+\dfrac{1}{x} \right]=2x-y\cos \left( xy \right)+\dfrac{y}{{{x}^{2}}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{2x-y\cos \left( xy \right)+\dfrac{y}{{{x}^{2}}}}{\left[ \cos \left( xy \right)+1+\dfrac{1}{x} \right]} \\\ \end{aligned}$$ Now, simplifying the above obtained equation we will get $$\Rightarrow \dfrac{dy}{dx}=\dfrac{2{{x}^{2}}-xy\cos \left( xy \right)+y}{x\cos \left( xy \right)+x+1}$$ Hence above is the required solution of the given equation. **So, the correct answer is “Option 4”.** **Note:** The possibility of mistake is that if students consider y as a constant and solve accordingly they will get the incorrect solution. Alternatively we can simplify the equation and convert it into the function of x alone and then differentiate the equation. The point to be remembered while solving the implicit differentiation is that we need to add $\dfrac{dy}{dx}$ every time we find a y term in the equation.