If \(xy^{2} = e^{y}\), where \(x \gt 0\), \(y \gt 0\) and \(y \ne 2\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{y}{x(y - 2)}\)
Differentiating both sides with respect to x (implicit differentiation, product rule on the left), with y′ for \(\frac{dy}{dx}\):
\(y^{2} + 2xyy' = e^{y}y'\)
\(y'\left(e^{y} - 2xy\right) = y^{2}\)
Now use the given relation \(e^{y} = xy^{2}\):
\(e^{y} - 2xy = xy^{2} - 2xy = xy(y - 2)\)
\(y' \cdot xy(y - 2) = y^{2}\)
\( \displaystyle y' = \frac{y^{2}}{xy(y - 2)} = \frac{y}{x(y - 2)} \)