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More derivatives!

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

A:  First we re-write:

f(x) = 1/√(x) = 1/x1/2 = x-1/2

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

In my words:  xa/b = “the b-th root” of xa ”  So, x1/2 is the “square-root of x1“….. x2/3 is the “cube-root of x2“… 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/x1/2 = 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

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