Answer to Question #106458 in Calculus for Nikesh

Question

Differentiate w.r.t x

y= √{x+√(x+√x)}

Answer 1

$y=\sqrt{x+\sqrt{x+\sqrt{x}}}\\ \text{We apply chain rule while differentiating}\\ \frac{dy}{dx}=\frac{d}{dx}(\sqrt{x+\sqrt{x+\sqrt{x}}})\\ \text{We know that }\frac{d}{dx}(\sqrt{g(x)})=\frac{1}{2\sqrt{g(x)}} \frac{d}{dx}(g(x))\\ \text{So, we have}\\ \frac{dy}{dx}=\frac{d}{dx}(\sqrt{x+\sqrt{x+\sqrt{x}}})\\ \;\;\;\;\;=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\frac{d}{dx}(x+\sqrt{x+\sqrt{x}})\\ \;\;\;\;\;=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\frac{1}{2\sqrt{x+\sqrt{x}}}\frac{d}{dx}(x+\sqrt{x}))\\ \;\;\;\;\;=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}}))\\$