# Differentiate the function by forming difference quotient [f(x+h)-f(x)]/h and taking the limit as h tends to 0. f(x)=Square Root(x-1)

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### 2 Answers

To calculate the value of the first derivative in a given point, h = 0, we'll have to apply the limit of the ratio:

limit [f(x+h)-f(x)]/h, when h tends to 0.

We'll calculate the value of f(x+h) and we'll substitute f(x+h):

f(x+h) = f(x+0) = f(x)

The limit will become:

limit [f(x+h)-f(x)]/h = limit [f(x)-f(x)]/h , h -> 0

**limit [f(x)-f(x)]/h = 0/0 indetermination case**

We'll apply L'Hospital rule. We'll differentiate separately both numerator and denominator, with respect to x.

[f(x)-f(x)]' = f'(x) - f'(x)

f'(x) = [sqrt(x-1)]'

f'(x) = (x-1)'/ 2sqrt(x-1)

f'(x) = 1/2sqrt(x-1)

f'(x) - f'(x) = 1/2sqrt(x-1) - 1/2sqrt(x-1) = 0

We'll differentiate the denominator h(x) = (x-0)

h'(x) = x'

h'(x) = 1

limit [f(x)-f(x)]'/h'(x) = 0/1

**limit [f(x+h)-f(x)]'/h'(x) = 0**

f(x) = sqrt(x-1).

To find the differential coefficient by taking the lt x-->0 {f(x+h)-f(x)}/h.

Given f(x) = sqrt(x-1)

Therefore , f'(x) = lt x -->0{f(x+h) -f(x)}/h

f'(x) = lt sqrt{(x+h-1) - sqrt(x-1)}/h '

We rationalise numerator, by multiplying both numerator and denominator by {sqrt(x+h)-sqrt(x-1)}:

f'(x) = lt {sqrt {x+h-1)-sqrt(x-1)}{(sqrt(x+h-1)+sqrt(x-1)}/h(sqrt(x+h-1)-sqrt(x-1)

f'(x) = lt {(x+h-1) -(x-1)/h{sqrt(x+h-1)+sqt(x-1)}

f'(x) = lt (h/h)/{sqrt(x+h-1)+swrt(x-1)}

f'(x) = 1/sqrt{x+0-1+sqrt(x-1)}

f'(x) = 1/{2sqrt(x-1)}