4.5.2 The Derivative of the Square Root Function

Let's write `y=sqrt(x)`. The difference quotient that approximates `dy text[/] dx` is

It's not obvious how to simplify this expression — in particular, how to find a factor of `Delta x` in the numerator. We need to use algebra to get a more useful form — but the most useful manipulation might not occur to you immediately.

The difference of square roots is the source of the difficulty. We could make that go away if we rationalize the numerator. It's the same idea you learned in high school for rationalizing a denominator: Multiply and divide by the sum of the square roots. Since we are multiplying and dividing by the same quantity, we are not changing the value of the expression. Let's see what happens:

Activity 1

1. Give a reason for each step in this calculation.

2. Finish the calculation of `dy text[/] dx` by determining the limiting value as `Delta x` approaches `0`.

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Example 1

Calculate the derivative of `sqrt(x) e^x`.

Solution   By the Product Rule,

Now, using our new formula for the derivative of the square root function, we find

Activity 2

1. Calculate the derivative of `xsqrt(x)` by using a difference quotient and a limiting value, as in the calculation preceding Activity 1.

2. Calculate the same derivative by using the Product Rule and your formula for the derivative of the square root function.

3. Compare the results of parts (a) and (b), both in terms of algebraic equivalence of the two answers and in terms of labor required to get the answers.

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