# Find the derivative:Find the derivative of y = sqrt x, from the first principles.

justaguide | Certified Educator

The question requires finding the derivative of y = sqrt x from the first principle. It isn't possible to use l"Hopital's rule here as that uses the derivative of what we are trying to find directly.

For y = sqrt x

y' = lim h -->0 [ (sqrt (x + h) - sqrt x)/h]

=> lim h -->0 [ (sqrt (x + h) - sqrt x)/(x + h - x)]

=> lim h -->0 [ (sqrt (x + h) - sqrt x)/((sqrt (x + h))^2 - (sqrt x)^2)]

=> lim h -->0 [ (sqrt (x + h) - sqrt x)/((sqrt (x + h) - sqrt x)((sqrt (x + h) + sqrt x)]

=> lim h -->0 [ 1/((sqrt (x + h) + sqrt x)]

substitute h = 0

=> 1/(sqrt x + sqrt x)

=> (1/2)*(1/sqrt x)

The derivative of y = sqrt x is y' = (1/2)*(1/sqrt x)

giorgiana1976 | Student

We'll put f(x) = y = sqrt x

f'(x) = lim [f(x) - f(0)]/(x-0), for x->0

f'(x) = lim (sqrtx - sqrt0)/x

We'll substitute x by 0:

lim (sqrtx - sqrt0)/x = (sqrt0 - sqrt0)/0 = 0/0

Since we have an indetermination case, 0/0, we'll apply L'Hospital rule:

lim f/g = lim f'/g'

lim sqrtx/x = lim (sqrt x)'/x'

lim (sqrt x)'/x' = lim (1/2sqrt x)/1

We'll substitute x by 0 and we'll get:

lim (1/2sqrt x)/1 = (1/2sqrt 0)

f'(x) = 1/2 sqrt x

f'(0) = 1/2