10. If \( f(x) = \sqrt{x + \sqrt{x}} \), then \( f'(x) = \)






Answer is: option4

\( \frac{2\sqrt{x} + 1}{4\sqrt{x^2 + x\sqrt{x}}} \)

Solution:

\( f(x) = \sqrt{x + \sqrt{x}} \)

\( f'(x) = \frac{1}{2 \sqrt{x + \sqrt{x}}} \cdot \left(1 + \frac{1}{2 \sqrt{x}} \right) \)

\( f'(x) = \frac{1}{2 \sqrt{x + \sqrt{x}}} \cdot \left(\frac{2 \sqrt{x} + 1}{2 \sqrt{x}} \right) \)

\( f'(x) = \frac{2 \sqrt{x} + 1}{4 \sqrt{x} \sqrt{x + \sqrt{x}}} \)

\( f'(x) = \frac{2 \sqrt{x} + 1}{4 \sqrt{x^2 + x \sqrt{x}}} \)Ans

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