If \(y = x^{\sqrt{x}}\), where \(x \gt 0\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{x^{\sqrt{x}}(2 + \log_e x)}{2\sqrt{x}}\)
The variable is in both the base and the exponent, so we use logarithmic differentiation:
\(\log_e y = \sqrt{x}\log_e x\)
By the product rule:
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \frac{\log_e x}{2\sqrt{x}} + \sqrt{x} \cdot \frac{1}{x} \)
\( \displaystyle = \frac{\log_e x}{2\sqrt{x}} + \frac{2}{2\sqrt{x}} = \frac{2 + \log_e x}{2\sqrt{x}} \)
\( \displaystyle \frac{dy}{dx} = \frac{x^{\sqrt{x}}(2 + \log_e x)}{2\sqrt{x}} \)
Why the other options are incorrect
Option (1) \(\sqrt{x}\,x^{\sqrt{x} - 1}\) — Incorrect. This treats the exponent \(\sqrt{x}\) as a constant (power rule), which is not allowed when the exponent varies.
Option (2) \(\frac{x^{\sqrt{x}}(1 + \log_e x)}{2\sqrt{x}}\) — Incorrect. The term \(\sqrt{x} \cdot \frac{1}{x} = \frac{1}{\sqrt{x}}\) equals \(\frac{2}{2\sqrt{x}}\), not \(\frac{1}{2\sqrt{x}}\).
Option (3) \(\frac{x^{\sqrt{x}}(2 + \log_e x)}{\sqrt{x}}\) — Incorrect. The factor \(\frac{1}{2}\) from the derivative of \(\sqrt{x}\) is lost; this is twice the correct value.