If \(y = (1 + x)^{x}\), \(x \gt 0\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(x = 1\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1 + 2\log_e 2\)
Since both the base and the power contain x, use logarithmic differentiation:
\(\log_e y = x\log_e(1 + x)\)
Differentiating with respect to x:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \log_e(1 + x) + \frac{x}{1 + x} \)
\( \displaystyle \frac{dy}{dx} = (1 + x)^{x}\left[\log_e(1 + x) + \frac{x}{1 + x}\right] \)
At x = 1, y = 2:
\( \displaystyle \frac{dy}{dx} = 2\left[\log_e 2 + \frac{1}{2}\right] \)
\( = 1 + 2\log_e 2 \)