Question

Differentiate the function w.r.t.x. : $\sqrt{x + \sqrt{x + \sqrt{x}}}$

Answer 1

really very good question.

Answer 2

Answer:

$\frac{d}{dx} \sqrt{x+\sqrt{x+\sqrt{x} } } =\frac{(2\sqrt{x+\sqrt{x} }+1)(2\sqrt{x} +1 )}{8x(\sqrt{x+\sqrt{x} } )\sqrt{x+\sqrt{x+\sqrt{x} } }}$

Step-by-step explanation:

As we know that

$\frac{d\sqrt{x} }{dx}=\frac{1}{2\sqrt{x} }$

so by this way

$\frac{d}{dx} \sqrt{x+\sqrt{x+\sqrt{x} } } =\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x} } }} .\frac{d}{dx} (x+\sqrt{x+\sqrt{x} })\\ \\ \\= \frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x} } }} . (1+\frac{ 1}{2\sqrt{x+\sqrt{x} })}).\frac{d}{dx}(x+\sqrt{x})\\ \\ \\=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x} } }} . (1+\frac{ 1}{2\sqrt{x+\sqrt{x} })}).(1+\frac{1}{2\sqrt{x}}). \frac{d\sqrt{x} }{dx}$

$=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x} } }} . (1+\frac{ 1}{2\sqrt{x+\sqrt{x} })}).(1+\frac{1}{2\sqrt{x}}). \frac{1}{2\sqrt{x}}$

Now simplify to get the answer

$\frac{d}{dx} \sqrt{x+\sqrt{x+\sqrt{x} } } =\frac{(2\sqrt{x+\sqrt{x} }+1)(2\sqrt{x} +1 )}{8x(\sqrt{x+\sqrt{x} } )\sqrt{x+\sqrt{x+\sqrt{x} } }}$