Let f(x) = xsqrt{xsqrt{xsqrt{xdotsinfty}}} wh | vedclass

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(B) Given $f(x) = x\sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$.Dividing by $x$,we get $\frac{f(x)}{x} = \sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$.Since the expres

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$6$

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(B) Given $f(x) = x\sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$.Dividing by $x$,we get $\frac{f(x)}{x} = \sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$.Since the expression under the square root is the same as the original function,we have $\frac{f(x)}{x} = \sqrt{f(x)}$.Squaring both sides,we get $\frac{f(x)^2}{x^2} = f(x)$.Assuming $f(x) 
eq 0$,we divide by $f(x)$ to get $\frac{f(x)}{x^2} = 1$,which implies $f(x) = x^2$.Now,differentiating with respect to $x$,we get $f'(x) = 2x$.Substituting $x = 3$,we get $f'(3) = 2(3) = 6$.

Let $f(x) = x\sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$ where $x > 0$. Then $f'(3)$ is equal to:

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