Use the definition of the derivititve to find f(x)= x + sqrt (x)?

Question

How do i use the definition of the derivititve to find f(x)= x + sqrt (x)

Answer 1

The definition of the derivative is
f'(x) = lim(1/h){f(x + h) -f(x)}
h->0
f'(x) =(1/h){ (x + h) + √(x + h) - [x + √x]}

f'(x) = (1/h){ h + √(x + h) - √x}
f'(x) = 1 + [√(x + h)] -√x]/ h
Multiply top and bottom of second term by [√(x + h)] +√x]
f'(x) = 1 + [√(x + h)] -√x][√(x + h)] +√x]/ h[√(x + h)] +√x]
f'(x) = 1 + [x + h - x]/h[√(x + h)] +√x]
f'(x) = 1 + h/h[√(x + h)] +√x]
f'(x) = 1 + 1/[√(x + h)] +√x]
but remember this is a limit as h--> 0
f'(x) = 1 + 1/(2√x)

Answer 2

Since you asked for a computation straight from the definition:

[h→0]lim (f(x+h) - f(x))/h
[h→0]lim ((x+h + √(x+h)) - (x + √x))/h
[h→0]lim (h + √(x+h) - √x)/h
[h→0]lim h/h + (√(x+h) - √x)/h
[h→0]lim 1 + (x+h - x)/(h(√(x+h) + √x))
[h→0]lim 1 + h/(h(√(x+h) + √x))
[h→0]lim 1 + 1/(√(x+h) + √x)
1 + 1/(√(x+0) + √x)
1 + 1/(2√x)

Answer 3

sqrt (x) = x^(1/2)

f(x) = x + x^(1/2)
f'(x) = 1 + (1/2)x^(-1/2)