If \( xy\,\log_e(x+y)=1 \), then \( \displaystyle\frac{dy}{dx}= \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{-(x+y+x^2y)}{x^2[y+(x+y)\,\log_e(x+y)]} \)
Differentiating implicitly, with the product rule on the three factors x, y and \( \log_e(x+y) \):
\( \displaystyle y\log_e(x+y)+x\log_e(x+y)\frac{dy}{dx}+\frac{xy}{x+y}\left(1+\frac{dy}{dx}\right)=0 \)
Collecting the \( \frac{dy}{dx} \) terms:
\( \displaystyle \frac{dy}{dx}\left[x\log_e(x+y)+\frac{xy}{x+y}\right]=-\left[y\log_e(x+y)+\frac{xy}{x+y}\right] \)
Multiplying both sides by (x + y):
\( \displaystyle \frac{dy}{dx}\cdot x\left[(x+y)\log_e(x+y)+y\right]=-\left[y(x+y)\log_e(x+y)+xy\right] \)
On the right side, the given relation gives \( \log_e(x+y)=\frac{1}{xy} \):
\( \displaystyle y(x+y)\cdot\frac{1}{xy}+xy=\frac{x+y}{x}+xy \)
\( \displaystyle =\frac{x+y+x^2y}{x} \)
Hence
\( \displaystyle \frac{dy}{dx}=-\frac{x+y+x^2y}{x\cdot x\left[y+(x+y)\log_e(x+y)\right]} \)
\( \displaystyle =\frac{-(x+y+x^2y)}{x^2[y+(x+y)\log_e(x+y)]} \)