How can the power rule be used to differentiate fractions in a simpler way?

[DorumonSg]

Its more a simplifying problem...

I was trying to differentiate this using definition principal

1-x^1/2

But I got stuck here :

(1-(x+h)^1/2 - (1-x^1/2))/h

I mean how do you explan something to the power of half or infact any fraction? I know I can change it to 1/sqrt(x+h)... but it just makes things more complicated... coz I can't get rid of the h then...

But using power rule is simple... 1-x^1/2 = -1/2X^-1/2

On top of that, I want to ask... are the constant numbers ignorable in differentiation? For example 4 - x^2 is -2x, I can just apply power rule on -x^2 and get the same answer... If I were to differentiate this by definition, can I just do -x^2 and ignore 4 too?

 

[lurflurf]

write sqrt(x+h)-sqrt(x) as
[(x+h)-x]/[sqrt(x+h)+sqrt(x)]
by using
a-b=(a^2-b^2)/(a+b)

We do not ignor constants, but differentiation is linear that is
(a f+b g)'=a f'+b g'
when a and b are constants and f and g are functions

 

[DorumonSg]

write sqrt(x+h)-sqrt(x) as
[(x+h)-x]/[sqrt(x+h)+sqrt(x)]
by using
a-b=(a^2-b^2)/(a+b)

Erm... so I am guessing you canceled the 1 away already before coming to this step? But if I do this its a fraction over h...

Coz the equation will become {[(x+h)-x]/[sqrt(x+h)+sqrt(x)]}/h and I still don't know how to get rid of h...

We do not ignor constants, but differentiation is linear that is
(a f+b g)'=a f'+b g'
when a and b are constants and f and g are functions

Erm... don't understand. Well I just did a differentiation using definition just now, and I realized you cannot remove the constant beforehand but when using Power Rule, it seems it can be ignored.

And how do you tell if it linear anyway?

 

[Bohrok]

What does (x+h)-x simplify to? After that, can you cancel the h?

 

[lurflurf]

-yes I canceled the one
-the h's cancel after the x's cancel in (x+h)-x
-one can prove the differentiation is linear or take it as an axiom
have you seen limit theorems like
lim[a f(x)]=a lim f(x)
and
lim [f(x)+g(x)]=lim f(x)+lim g(x)
if so it will be easy to show
(a f+b g)'=a f'+b g'
by using the limit definition of differentiation
-can you finish the differentiation of sqrt(x)?