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

Lcm of 6 and 8 is 24lcm of 4 and 12 is 12lcm of 11 and 12 is 132 lcm of 3 and 6 is 6