determine where f is differentiable..

**tn11631** · Apr 18th 2010, 11:23 AM

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

**southprkfan1** · Apr 18th 2010, 12:29 PM

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!

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

Odds are I screwed up somewhere

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