On-line Math 21

2.1  The definition of the derivative

Example 4 Find f¢(x) for the function f(x) = Öx .

Solution

The next step is to multiply top and bottom by

(

___ Öx+h

+Öx)

.

f¢(x) = lim h® 0  f(x+h)-f(x) h = lim h® 0    ___ Öx+h   -Öx h = lim h® 0  æ è   ___ Öx+h   -Öx ö ø æ è   ___ Öx+h   +Öx ö ø h æ è   ___ Öx+h   +Öx ö ø .

This trick is known as conjugation. Any time we have a difference of square roots, we can simplify that by multiplying, numerator and denomonator, by the sum of the square roots (the conjugate). The idea, contrary to what you were probably taught before, is to rationalize the numerator, not the denominator. We can do more with the numerator algebraically once we rationalize it than we could with the denominator. I never have understood why you are supposed to rationalize the denominator in fractions. I think that

1 Ö2

makes more sense than

Ö2 2 ,

anyway.

Now that we have multiplied by this conjugate, what happens?

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Copyright (c) 2000 by David L. Johnson.