CUET Preparation Today
If $f(x)=\sqrt{1+\sqrt{x}}$ for x > 0, then f'(x) is : |
$\frac{1}{2 f(x)}$ $\frac{1}{4 \sqrt{x} f(x)}$ $\frac{2 \sqrt{x}+1}{4 \sqrt{x} f(x)}$ $\frac{1}{2 \sqrt{x} f(x)}$ |
$\frac{1}{4 \sqrt{x} f(x)}$ |
$f'(x)=\frac{1}{2 \sqrt{1+\sqrt{x}}}\left(0+\frac{1}{2 \sqrt{x}}\right)$ $=\frac{1}{4 \sqrt{x}(\sqrt{1+\sqrt{x}})}=\frac{1}{4 \sqrt{x} . f(x)}$ Hence (2) is correct answer. |
