Chapter 2
Models of Growth: Rates of Change





2.3 Symbolic Calculation of Derivatives: Polynomial Functions

2.3.2 Derivatives of Power Functions: General Rules

Based on the examples so far, we may formulate a general rule for differentiating power functions of the form t n .

Power Rule   If n is any positive integer, then
d d t t n = n t n - 1 .

Next we consider derivatives of functions such as f ( t ) = 6 t 3 . Recall that we have already shown that

d d t c t 2 = 2 c t = c d d t t 2 .

For general derivatives of this form, we need the Constant Multiple Rule.


Constant Multiple Rule   If f ( t ) is any function that has a derivative, and \(c\) is any constant, then
d d t c f ( t ) = c d f d t .

That is, constant factors can be factored out of derivative calculations.

Example 1   Use the Constant Multiple Rule to calculate the derivative of 6 t 3 .

Solution
d d t 6 t 3 = 6 d d t t 3 = 6 ( 3 t 2 ) = 18 t 2

 

The Constant Multiple Rule follows from some simple algebra with difference quotients. If s = c f ( t ) , then

Δ s Δ t = c f ( t + Δ t ) - c f ( t ) Δ t
= c f ( t + Δ t ) - f ( t ) Δ t .

As Δ t 0 , we obtain

d d t c f ( t ) = c d f d t .

We need one more differentiation rule to be able to differentiate general polynomials: The Sum Rule. In words, the Sum Rule says that the derivative of the sum of two functions is the sum of the derivatives of the two functions. This can be established in much the same way as the Constant Multiple Rule.

Sum Rule   If f ( t ) and g ( t ) are any functions that have derivatives, then
d d t [ f ( t ) + g ( t ) ] = d f d t + d g d t .

Polynomials are just sums of terms of the form k t n for various values of \(k\) and \(n\). We now know how to differentiate each such term, and we also know how to differentiate sums.

Example 2   Apply the Sum, Constant Multiple, and Power Rules to differentiate the polynomial 3 t 5 - 2 t 3 + 7 t 2 + 3 .

Solution
d d t ( 3 t 5 - 2 t 3 + 7 t 2 + 3 ) = d d t ( 3 t 5 ) + d d t ( - 2 t 3 ) + d d t ( 7 t 2 ) + d d t ( 3 )
= 3 d d t t 5 - 2 d d t t 3 + 7 d d t t 2 + 0
= 3 5 t 4 - 2 3 t 2 + 7 2 t
= 15 t 4 - 6 t 2 + 14 t

 

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