Using the limit definition, how do you find the derivative of f (x) = sqrt (x 1) ?

= √x

y=x12

Differentiate w. r. t "x" on both sides:

dydx=ddx[x12]

dydx=12x12-1 (because ddx[xn]=nxn-1)

dydx=12x-12

And it can also be written as:

dydx=12√x

Or, if you meant the limit definition of the derivative function it would look like this:

f' (x) = limh→0f (x+h) - f (x) h

f' (x) = limh→0√x+h-√xh

Now, we multiply the numerator and the denominator by the conjugate of the numerator (conjugates are the sum and difference of the same two terms such as a + b and a - b).

f' (x) = limh→0√x+h-√xh⋅√x+h+√x√x+h+√x

Since (a+b) (a-b) = a2-b2 we get

f' (x) = limh→0x+h-xh (√x+h+√x)

Simplifying, we get

f' (x) = limh→0hh (√x+h+√x)

f' (x) = limh→01√x+h+√x

If we evaluate the limit by plugging in 0 for h we get

f' (x) = 1√x+0+√x=1√x+√x=12√x