Question

If$S\mathrm{in}\left(xy\right)+\frac{x}{y}=x^2-y$ then $\frac{dy}{dx}$is equal to

• A. $\frac{y(2xy-x^2\cos xy-1)}{(x{y}^2\cos xy–x+y^2)}$
• B. $\frac{(2xy-y^2\cos xy-1)}{(x{y}^2\cos xy–x+y^2)}$
• C. $\frac{y(2xy-y^2\cos xy-1)}{(x{y}^2\cos xy–x+y^2)}$
• D. None of these

Answer 1

The correct option is C

$\frac{y(2xy-y^2\cos xy-1)}{(x{y}^2\cos xy–x+y^2)}$

Explanation for the correct option

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

Given information

$\Rightarrow \sin \left(xy\right)+\frac{x}{y}=x^2-y$

$\therefore \cos \left(xy\right)\left(\frac{d}{dx}\right)\left(xy\right)+\frac{\left(y–x{\displaystyle \frac{dy}{dx}}\right)}{y^2}=2x–\frac{dy}{dx}$

$\Rightarrow \cos \left(xy\right)\left(y+x\frac{dy}{dx}\right)+\frac{\left(y–x{\displaystyle \frac{dy}{dx}}\right)}{y^2}=2x–\frac{dy}{dx}$

$\Rightarrow y\cos \left(xy\right)+x\cos \left(xy\right)\frac{dy}{dx}+\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}=2x–\frac{dy}{dx}$

Step 2: Compare the above equation on both sides and take the differential on one side

$\Rightarrow x\cos \left(xy\right)\frac{dy}{dx}-\frac{x}{y^2}\frac{dy}{dx}+\frac{dy}{dx}=2x–y\cos \left(xy\right)-\frac{1}{y}$

$\Rightarrow \left(x\cos \left(xy\right)-\frac{x}{y^2}+1\right)\frac{dy}{dx}=2x–y\cos \left(xy\right)-\frac{1}{y}$

$\Rightarrow \frac{dy}{dx}=\frac{\left(2x–y\cos \left(xy\right)-\frac{1}{y}\right)}{\left(x\cos \left(xy\right)-\frac{x}{y^2}+1\right)}$

$\Rightarrow \frac{dy}{dx}=\frac{{\displaystyle \frac{1}{y}}\left(2xy–y^2\cos \left(xy\right)-1\right)}{{\displaystyle \frac{1}{y^2}}\left(x{y}^2\cos \left(xy\right)-x+y^2\right)}$

$\Rightarrow \frac{dy}{dx}=\frac{{\displaystyle y}\left(2xy–y^2\cos \left(xy\right)-1\right)}{\left(x{y}^2\cos \left(xy\right)-x+y^2\right)}$

Hence option (C) is the correct answer.