y= ##sqrt x##

y=##x^(1/2)##

Differentiate w.r.t “x” on both sides:

##dy/dx= d/dx [x^(1/2)]##

##dy/dx= 1/2x^(1/2-1)## (because ##d/dx[x^n]=nx^(n-1)##)

##dy/dx= 1/2x^(-1/2)##

And it can also be written as:

##dy/dx= 1/(2sqrt(x))##

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

##f'(x) = lim_(h->0)(f(x+h)-f(x))/h##

##f'(x) = lim_(h->0)(sqrt(x+h)-sqrtx)/h##

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) = lim_(h->0)(sqrt(x+h)-sqrtx)/h*(sqrt(x+h)+sqrtx)/(sqrt(x+h)+sqrtx)##

Since ## (a+b)(a-b) = a^2-b^2## we get

##f'(x)=lim_(h->0)(x+h-x)/(h(sqrt(x+h)+sqrtx)##

Simplifying, we get

##f'(x)=lim_(h->0)h/(h(sqrt(x+h)+sqrtx)##

##f'(x)=lim_(h->0)1/(sqrt(x+h)+sqrtx)##

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

##f'(x)=1/(sqrt(x+0)+sqrtx)=1/(sqrtx+sqrtx)=1/(2sqrtx)##