Find d/dx √x using the rule f(x+h) - f(x)/h with f(x) = √x

Find d/dx √x using the rule
f'(x) = lim h->0 f(x+h) - f(x)/h with f(x) = √x


I already know that d/dx √x is 1/2√x but how do i use the formula to figure it out.

Please help.

Yup, that's the definition of a derivative.

Given:
lim(h→0) [(√(x + h) − √x)/h]

Multiply by (√(x + h) + √x)/(√(x + h) + √x):
= lim(h→0) [(√(x + h) − √x)(√(x + h) + √x)/(h(√(x + h) + √x))]
= lim(h→0) [(x + h − x)/(h(√(x + h) + √x))]
= lim(h→0) [h/(h(√(x + h) + √x))]

The h in the numerator and denominator cancel out:
= lim(h→0) [1/(√(x + h) + √x)]

Let h approach zero:
= [1/(√(x + 0) + √x)]
= 1/(√x + √x)
= 1/(2√x)

f(x) = sqrt(x)
f(x + h) = sqrt(x + h)

(f(x + h) - f(x)) / h =>
(sqrt(x + h) - sqrt(x)) / h

Now, the tricky part. Rationalize the numerator:

((sqrt(x + h) - sqrt(x)) * (sqrt(x + h) + sqrt(x))) / (h * (sqrt(x + h) + sqrt(x)) =>
(x + h - x) / (h * (sqrt(x + h) + sqrt(x)) =>
h / (h * sqrt(x + h) + sqrt(x)) =>
1 / (sqrt(x + h) + sqrt(x))

h goes to 0

1 / (sqrt(x + 0) + sqrt(x)) =>
1 / (sqrt(x) + sqrt(x)) =>
1 / (2 * sqrt(x))

lim (h->0)(f(x+h) - f(x))/h =

lim (h->0)(√x+h - √x)/h =

lim (h->0)((x+h-x)/(h(√x+h + √x)) =

lim (h->0)(1/(√x+h + √x)) = 1/2√x