Discussion

If \(e^{y} + x^{2}y + xy^{2} = e^{2}\), then \(\frac{dy}{dx}\) at (0,1) is equal to

Answer: E

\(\frac{dy}{dx}\) at (0, 1) is \(\frac{-1}{e}\).

Differentiate implicitly:

\[ e^{y}y' + 2xy + x^{2}y' + y^{2} + 2xyy' = 0 \]

At (0, 1) every term with a factor x vanishes, leaving \(e\,y' + 1 = 0\), so \(y' = -\frac{1}{e}\).

\(\frac{1}{e}\) drops the sign, and ±e invert the factor.

No comment is present. Be the first to comment.

Loading…