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emcrazy14Best ResponseYou've already chosen the best response.0
\[\frac{ 1+x }{ \sqrt{1x} }\]
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
expand in a power series i take it
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
about zero i hope first term is \(f(0)=1\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
first derivative is \[f'(x)=\frac{3x}{2\sqrt{1x}^3}\] so second term is \[f'(0)x=\frac{3}{2}x\]
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
yikes, now we need the derivative of that beast
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
do i have to do that too?
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
looks like it is \[f''(x)=\frac{7x}{4\sqrt{1x}^5}\]
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
making the next term \[\frac{f''(0)}{2}x^2=\frac{7}{8}x^2\]
 7 months ago

emcrazy14Best ResponseYou've already chosen the best response.0
So I find the derivative of that too?
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
\[f'^{(3)}(x)=\frac{333x}{8\sqrt{1x}^7}\]
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
if so then \(\frac{f^{(3)}(0)}{3!}x^3=\frac{33}{16}x^3\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
hmm let me check if this is right
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
no messed up \(3!=6\) doh
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
third term is \(\frac{11}{16}x^3\)
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
k lets check the whole dam thing and see if it is right
 7 months ago

satellite73Best ResponseYou've already chosen the best response.2
http://www.wolframalpha.com/input/?i=%281%2Bx%29%2Fsqrt%281x%29 yup !!
 7 months ago

emcrazy14Best ResponseYou've already chosen the best response.0
Okay, thank you so much! :)
 7 months ago
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