If \(xy+\displaystyle\frac{x^2}{y}=x^3y+y\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{y(3x^2y^2-2x-y^2)}{xy^2-x^3y^2- x^2 - y^2}\)
Let \(y' = \frac{dy}{dx}\). Differentiating both sides with respect to x (implicit differentiation), using the product rule for \(xy\) and \(x^3y\) and the quotient rule for \(\frac{x^2}{y}\):
\( \displaystyle y + xy' + \frac{2xy - x^2y'}{y^2} = 3x^2y + x^3y' + y' \)
Multiplying every term by \(y^2\):
\(y^3 + xy^2y' + 2xy - x^2y' = 3x^2y^3 + x^3y^2y' + y^2y'\)
Collecting the \(y'\) terms on the left:
\(y'(xy^2 - x^2 - x^3y^2 - y^2) = 3x^2y^3 - y^3 - 2xy\)
\(y'(xy^2 - x^3y^2 - x^2 - y^2) = y(3x^2y^2 - 2x - y^2)\)
\( \displaystyle y' = \frac{y(3x^2y^2 - 2x - y^2)}{xy^2 - x^3y^2 - x^2 - y^2} \)