If \(y = \log_e(x + y)\), where \(x + y \gt 0\), then the value of \(\displaystyle\frac{dy}{dx}\) at the point \((e - 1, 1)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{e - 1}\)
The point lies on the curve, since \(\log_e(e - 1 + 1) = \log_e e = 1\).
Differentiating both sides with respect to x (implicit differentiation), with y′ for \(\frac{dy}{dx}\):
\( \displaystyle y' = \frac{1 + y'}{x + y} \)
\(y'(x + y) = 1 + y'\)
\(y'(x + y - 1) = 1\)
\( \displaystyle y' = \frac{1}{x + y - 1} \)
At \((e - 1, 1)\), \(x + y = e\):
\( \displaystyle \frac{dy}{dx} = \frac{1}{e - 1} \)