= √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
