Chapter 4
Differential Calculus and Its Uses





4.7 The General Power Rule

4.7.1 Special Cases

We need one more general rule for differentiation. We know how to differentiate power functions such as `ftext[(]xtext[)]=x^2` and `gtext[(]xtext[)]=x^9` — the derivatives are `f'text[(]xtext[)]=2x` and `g'text[(]xtext[)]=9x^8`. We can summarize what we know about derivatives of power functions:

d d x x n = n x n - 1 .

Here, `n` can be `0` or `1` or `2` or any nonnegative integer. The cases of `n=1text[/]2` and `n=-1`, both discussed earlier in this chapter, also fit this formula:

d d x x 1 2 = d d x x = 1 2 x = 1 2 1 x = 1 2 x - 1 2 ,
 
d d x x - 1 = d d x 1 x = - 1 x 2 = ( - 1 ) x - 2 .

What about the derivative of `y=x^(7//3)`? Does it fit the form of the Power Rule? To answer this, we use implicit differentiation again. The `7text[/]3` power function satisfies the equation

y 3 = x 7 .

Now we can differentiate both sides with respect to `x` — because both exponents are positive integers:

3 y 2 d y d x = 7 x 6 .

When we solve for `dytext[/]dx`, we find

d y d x = 7 x 6 3 y 2 .

If we substitute `y=x^(7//3)`, we get

d y d x = 7 x 6 3 ( x 7 3 ) 2

or

d y d x = 7 3 x 6 - 14 3 = 7 3 x 4 3 .

Note that if we set `n=7text[/]3`, this fits the pattern we have already seen:

d d x x n = n x n - 1 .

This Power Rule formula holds for any rational power `n`. Indeed, it holds for any power function, whether `n` is rational or not, although it may not be clear at this point what an irrational power means.

Before we discuss how to establish a general Power Rule, we give another example that shows how we may use the Power Rule and the Chain Rule in combination to calculate a rather complicated derivative. This example is much like the calculations we did in Section 4.5 when we examined the reflection property of light.


Example 1

Differentiate the function `(e^(3x)+5x)^(7//3)`.

Solution   Using both the Power Rule and the Chain Rule, we find

d d x ( e 3 x + 5 x ) 7 3 = 7 3 ( e 3 x + 5 x ) 4 3 d d x ( e 3 x + 5 x ) .

When we finish calculating the second factor, we have

d d x ( e 3 x + 5 x ) 7 3 = 7 3 ( e 3 x + 5 x ) 4 3 ( 3 e 3 x + 5 ) .
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