bbafung

d/dx (1/√x)   

1 over 開方x

THANKS!!

[ 本帖最後由 bbafung 於 2014-6-16 11:04 PM 編輯 ]

samwong001

d/dx (1/√x)   
= d/dx x^(-1/2 )
=(-1/2) x^(-3/2)
=-1/(2x√x)

btw, it is called differentiation

[ 本帖最後由 samwong001 於 2014-6-17 10:11 AM 編輯 ]

bbafung

how to do by the first principles??

samwong001

d/dx (1/√x)   
= lim h-->0  ((1/√(x+h)) - 1/√(x)) / h        
= lim h-->0  √(x) -√(x+h)/ h√(x+h) √(x)   times √(x+h) √(x) to both numerator and denominator
= lim h-->0  (√(x) -√(x+h)) (√(x) +√(x+h)) / h√(x+h) √(x) (√(x) +√(x+h)  times  √(x) +√(x+h) to both numerator and denominator
= lim h-->0 -h/ h√(x+h) √(x) (√(x) +√(x+h)) simplify the numerator
= lim h-->0 -1/ 1√(x+h) √(x) (√(x) +√(x+h)) simplify the numerator
= -1/ √(x) √(x) (√(x) +√(x)) sub h=0
= -1/ x (2√(x))
=-1/(2x√x)

please check if there is any mistake~~ I may miss some ( or ) ....

[ 本帖最後由 samwong001 於 2014-6-17 10:11 AM 編輯 ]

kc3551@u-depot

last line should be = -1/(2x√x) ?

bbafung

thx so much ;)

samwong001

Thanks, I corrected it.