If $f(x)=\sqrt{1+\sqrt{x}}$ for x &

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Continuity and Differentiability

Question:

If $f(x)=\sqrt{1+\sqrt{x}}$ for x > 0, then f'(x) is :

Options:

$\frac{1}{2 f(x)}$

$\frac{1}{4 \sqrt{x} f(x)}$

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

$\frac{1}{2 \sqrt{x} f(x)}$

Correct Answer:

$\frac{1}{4 \sqrt{x} f(x)}$

Explanation:

$f'(x)=\frac{1}{2 \sqrt{1+\sqrt{x}}}\left(0+\frac{1}{2 \sqrt{x}}\right)$

$=\frac{1}{4 \sqrt{x}(\sqrt{1+\sqrt{x}})}=\frac{1}{4 \sqrt{x} . f(x)}$

Hence (2) is correct answer.