Question

If \(y=x^{\sqrt{x}}\), then \(\frac{dy}{dx}=\)

• A. \(\frac{y}{2\sqrt{x}}(2+\log x)\)
• B. \(\frac{y}{\sqrt{x}}(2+\log x)\)
• C. \(\frac{y}{2\sqrt{x}}(2-\log x)\)
• D. \(\frac{y}{\sqrt{x}}(2-\log x)\)

Hint:

Take the natural logarithm of both sides and differentiate implicitly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a variable power, take logarithms first: \(\log y=\sqrt{x}\,\log x\).

Step 2: Differentiate
\(\frac{1}{y}\frac{dy}{dx}=\sqrt{x}\cdot\frac{1}{x}+\log x\cdot\frac{1}{2\sqrt{x}}=\frac{1}{\sqrt{x}}+\frac{\log x}{2\sqrt{x}}\).

Step 3: Simplify
\(\frac{1}{y}\frac{dy}{dx}=\frac{2+\log x}{2\sqrt{x}}\), so \(\frac{dy}{dx}=\frac{y}{2\sqrt{x}}(2+\log x)\).

Final Answer:
The derivative is \(\frac{y}{2\sqrt{x}}(2+\log x)\). \[ \boxed{\text{(A) }\dfrac{y}{2\sqrt{x}}(2+\log x)} \]