If $y=\sqrt{x+\sqrt{x+\sqrt{x}}}, x>0$, then $d y / d x$ at $x=1$ is

$\frac{1}{\sqrt{1+\sqrt{2}}}\left(\frac{3+4 \sqrt{2}}{4 \sqrt{2}}\right)$

$\left(\frac{3+4 \sqrt{2}}{8 \sqrt{2}}\right)$

$\frac{3+4 \sqrt{2}}{\sqrt{1+\sqrt{2}}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y=\sqrt{x+\sqrt{x+\sqrt{x}}}, x>0

, then

d y / d x

x=1

View answer and explanation

Correct option: A

\frac{1}{2 \sqrt{1+\sqrt{2}}}\left(\frac{3+4 \sqrt{2}}{4 \sqrt{2}}\right)

Option A (

\frac{1}{2 \sqrt{1+\sqrt{2}}}(\frac{3+4 \sqrt{2}}{4 \sqrt{2}})

) is correct because it follows from applying the chain rule to find the derivative and then substituting x=1.

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