Thread: determine where f is differentiable..

Define f(x)=sqrt(x+sqrt(x+sqrt(x))) for x >=0. Determine where f is differentiable and compute the derivative.

Originally Posted by tn11631

Define f(x)=sqrt(x+sqrt(x+sqrt(x))) for x >=0. Determine where f is differentiable and compute the derivative.

Chain rule!

$\displaystyle f'(x) = \frac{1}{2(x+\sqrt{x+\sqrt{x}})^{1/2}}*[1+(\frac{1}{2(x+\sqrt{x})^{1/2}})*(1+\frac{1}{2x^{1/2}})] $

Odds are I screwed up somewhere