so why didn't you show your work?
f(x+h)-f(x) = 1/√(x+h) - 1/√x
Now divide by h and you have
(1/√(x+h) - 1/√x)/h
= (√x - √(x+h)) / (h(√x)(√x+h))
= (√x - √(x+h))(√x + √(x+h)) / (h √x √(x+h))
= (x - (x+h)) / (h √x √(x+h) * (√x + √(x+h)))
= -h / (h √x √(x+h) * (√x + √(x+h)))
= -1/ (√x √(x+h) * (√x + √(x+h)))
Now take the limit as h→0 and you have
-1/(√x √x * 2√x)
= -1/(2x^(3/2))
using first principles to differentiate y = 1/(√ x).
I tried using this method many times but I get different answers all the time. I can do it using the power rule but I am not sure what I am doing wrong....
my answer is -1/(2x^3/2) using power rule
2 answers
y = 1/(√ x) = x^-0.5
dy/ dx = -0.5 x^-1.5 so I agree with your answer
now
y ( x+h )= (x+h)^-.5 = 1/(x+h)^.5
y (x) = 1/x^.5
y (x+h) -y(x) = 1/(x+h)^.5 - 1/x^.5
= [ x^.5 - (x+h)^.5] / [x^.5 (x+h)^.5]
we want that divided by h as h --->0
well (a+b)^.5 = a^.5 + ,5 a^-.5 b +.5 *.5*-.5*a^-1.5 bb^2 .... series
so [ x^.5 - x^-5 - .5 x^-.5 h .....] / h [x^.5 (x+h)^.5]
let h ---> 0
-.5 x^-.5 / x
= -.5 x^-1.5
Caramba
dy/ dx = -0.5 x^-1.5 so I agree with your answer
now
y ( x+h )= (x+h)^-.5 = 1/(x+h)^.5
y (x) = 1/x^.5
y (x+h) -y(x) = 1/(x+h)^.5 - 1/x^.5
= [ x^.5 - (x+h)^.5] / [x^.5 (x+h)^.5]
we want that divided by h as h --->0
well (a+b)^.5 = a^.5 + ,5 a^-.5 b +.5 *.5*-.5*a^-1.5 bb^2 .... series
so [ x^.5 - x^-5 - .5 x^-.5 h .....] / h [x^.5 (x+h)^.5]
let h ---> 0
-.5 x^-.5 / x
= -.5 x^-1.5
Caramba