Question

find f'(x) given f(x)= (sqrt 1+ sqrt 1+x^2)

Answer 1

\[f(x)=\sqrt 1 + \sqrt {1+x^2}\]??

Answer 2

ooh maybe
\[f(x)=\sqrt{1+\sqrt{1+x^2}}\]

Answer 3

yes . I know the formula for f ' (x) = lim as z approaches x f(z) - f(x) / z-x correct?

Answer 4

oh yes

Answer 5

i wouldn't try that here, it would take all night

Answer 6

i would use the fact that
\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x))}}\]

Answer 7

ahhh

Answer 8

put \(f(x)=1+\sqrt{1+x^2}\) and \(f'(x)=\frac{x}{\sqrt{1+x^2}}\) in the above formula
\[

Answer 9

ok I will try that.