**Q: Find the derivative when x = 1 of the function f(x) = 1/√(x)**

A: First we re-write:

f(x) = 1/√(x) = 1/x^{1/2} = x^{-1/2}

How did I get that??? Let’s rewind a bit and talk some basics:

In my words: x^{a/b} = “the b-th root” of x^{a} ” So, x^{1/2} is the “square-root of x^{1}“….. x^{2/3} is the “cube-root of x^{2}“… Make sense? Therefore, I was able to turn the “square-root” sign into a 1/2 power.

And, a **negative exponent** puts you where you’re not. If you are in the denominator, a negative exponent can pull you up the the numerator. If you are in the numerator, a negative exponent puts you in the denominator. That is how I was able to “get rid” of the fraction and pull the x up… by adding a negative to the exponent!

OK. Long-winded explanation, but hopefully helpful… So, back to the problem:

f(x) = 1/√(x) = 1/x^{1/2} = x^{-1/2}

Now, take the derivative of f(x) = x^{-1/2} using the power-rule:

f(x) = x^{-1/2}

f ‘ (x) = (-1/2)x^{-1/2 – 1}

f ‘ (x) = (-1/2)x^{-3/2}

Now, to find f ‘ (1), plug in 1!

f ‘ (x) = (-1/2)(1)^{-3/2} = (-1/2)(1) = -1/2