# Calculus Help

posted by Mello

Use f(x)= 1/(square root(x)+1) to answer the following;
-- Use the difference quotient f'(x)=lim z->x (f(x)-f(z))/ (x-z) to find f'(x)!

Thank you!!!

1. Damon

f(x) = 1/[ x^.5 +1 ]
f(z) = 1/[ z^.5 + 1 }

f(x)-f(z) = 1/[ x^.5 +1 ] - 1/[ z^.5 + 1]

= ([z^.5+1]-[x^.5+1])/(x^.5z^.5 +x^.5+z^.5+1)

= (z^.5-x^.5) /(x^.5z^.5 +x^.5+z^.5+1)
Divide by (x-z) which is (x^.5-z^.5)(x^.5+z^.5)
and get
-(x^.5+z^.5) / (x^.5z^.5 +x^.5+z^.5+1)
let z --->x
- 2 x^.5 / (x +2 sqrt x + 1)
check my arithmetic !!!

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