If \(y = x^{x^{2}}\), where \(x \gt 0\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(x = e\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3e^{e^{2} + 1}\)
The variable is in both the base and the exponent, so we use logarithmic differentiation:
\(\log_e y = x^{2}\log_e x\)
By the product rule:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = 2x\log_e x + x^{2} \cdot \frac{1}{x} = 2x\log_e x + x \)
\( \displaystyle \frac{dy}{dx} = x^{x^{2}}\left(2x\log_e x + x\right) \)
At x = e, \(\log_e e = 1\) and \(y = e^{e^{2}}\):
\( \displaystyle \frac{dy}{dx} = e^{e^{2}}(2e + e) = 3e \cdot e^{e^{2}} = 3e^{e^{2} + 1} \)
Why the other options are incorrect
Option (1) \(2e^{e^{2} + 1}\) — Incorrect. This keeps only the term \(2x\log_e x\) of the product rule and drops the term x.
Option (2) \(e^{e^{2} + 1}\) — Incorrect. This keeps only the term x and drops \(2x\log_e x\).
Option (4) \(3e^{e^{2}}\) — Incorrect. The bracket at x = e is 3e, not 3, so a factor e is missing.