Differentiating y=x^-1/2 from first principles

Anyone know how to, or can provide a link to, how to differentiate y = x^(-1/2) from first principles?
I'm assuming it involves some kind of power series.

Thanks!

The first principle of the derivative is:

Lim {f(x + h) - f(x)}/h
h→0

let f(x) = x^(-1/2), then f(x + h) = (x + h)^(-1/2)

Lim {(x + h)^(-1/2) - x^(-1/2)}/h
h→0

Substitute 1/(x + h)^(1/2) for (x + h)^(-1/2) and 1/x^(1/2) for x^(-1/2):

Lim {1/(x + h)^(1/2) - 1/x^(1/2)}{1/h}
h→0

Add the two fractions using the common denominator h{x^(1/2)}(x + h)^(1/2):

Lim {x^(1/2) - (x + h)^(1/2)}/[h{x^(1/2)}(x + h)^(1/2)]
h→0

Multiply the numerator and denominator by {x^(1/2) + (x + h)^(1/2)}

Lim [{x^(1/2) - (x + h)^(1/2)}{x^(1/2) + (x + h)^(1/2)}]/[h{x^(1/2)}(x + h)^(1/2){x^(1/2) + (x + h)^(1/2)}]
h→0

Use the F.O.I.L. method to multiply the numerator:

{x^(1/2)}² - (x + h)^(1/2)(x^(1/2)) + (x + h)^(1/2)(x^(1/2)) - {(x + h)^(1/2)}²

Notice that the middle terms add up to zero and the end terms are the difference of squares of square roots:

x - (x + h) = -h

Notice that there is an h in the denominator that will leave us with a -1 in numerator:

Lim -1/[{x^(1/2)}(x + h)^(1/2){x^(1/2) + (x + h)^(1/2)}]
h→0

You may now let h go to zero:

Lim -1/[{x^(1/2)}(x + h)^(1/2){x^(1/2) + (x + h)^(1/2)}] = -1/[{x^(1/2)}(x)^(1/2){x^(1/2) + (x)^(1/2)}] =
h→0

(-1/2)1/x^(3/2) = (-1/2)x^(-3/2)

y = x^(-1/2) = 1/sqrtx = f(x)

f'(x) = dy/dx

dy/dx = lim (h->0) (f(x+h) - f(x))/h

= lim (h->0) (1/sqrt(x+h) + 1/sqrtx)/h

= lim (h->0) (sqrtx - sqrt(x+h))/hsqrtx*sqrt(x+h)

Multiply by conjugate of: sqrtx - sqrt(x+h)

dy/dx = lim (h->0) (sqrtx - sqrt(x+h))/hsqrtx*sqrt(x+h)*(sqrtx + sqrt(x+h))/(sqrtx + sqrt(x+h))

= lim (h->0) (x-x-h)/(h(xsqrt(x+h) +(x+h)sqrtx)

= lim (h->0) -h/h(xsqrt(x+h) +(x+h)sqrtx)

= lim (h->0) -1/(xsqrt(x+h) +(x+h)sqrtx)

As h->0, f'(x) -> -1/(xsqrtx + xsqrtx) = -1/(2xsqrtx) = (-1/2)x^(-3/2)

Differentiating y=x^-1/2

-(1/(2 x^(3/2)))

y = x^ (-1/2)

y' =( -1/2) x ^ (-1/2 -1)

y ' = - (1/2) x ^ (-3/2)

using : d/dx x^n = nx^(n-1)

: )