### Rules of differentiation: Rules of computation for the derivative

### The sum rule for differentiation

So far, we have been acquainted with the derivatives of a number of standard functions. In this chapter, we will look at some rules of calculation for dealing with sums, products, quotients, and composite functions of these standard functions.

We start with sums of functions. For example, consider the function #f(x)=4\cdot x^3+3 \cdot x#. With the help of two of the three rules in *Three basic rules for differentiation* we can determine the derivative of #f#. Here we give a more general rule that gives the derivative at once.

First, we define what we mean by a sum function.

Sum function

Let #f# and #g# be two functions and #a# and #b# two numbers. The **sum function** #a\cdot f+b\cdot g# is the function that assigns the value \[(a\cdot f+b\cdot g)(x) = a\cdot f(x) + b\cdot g(x)\] to #x#. Hence, #a\cdot f# is the function that assigns to #x# the value #(a\cdot f)(x) = a \cdot f(x)#.

For example, if #a =3# and #f(x) = x^2+1#, the function #a\cdot f# assigns the value #a\cdot f(x) = 3\cdot(x^2+1)=3x^2+3# to #x#.

In order to distinguish sums like #a\cdot f+b\cdot g# from the sum #f+g# we speak of **linear combinations** of #f# and #g#.

The extended sum rule for differentiation

Let #a# and #b# be constants and let #f# and #g# be differentiable functions. The derivative of #a\cdot f+b\cdot g# satisfies

\[ \left(a\cdot f+b\cdot g\right)' = a\cdot f'+b\cdot g'\tiny.\] The rule can be extended to sums of multiple functions. If, for example, #c# is a third constant and #h# a third differentiable function, then \[\left(a\cdot f+b\cdot g+c\cdot h\right)'=a\cdot f'+b\cdot g'+c\cdot h'\tiny.\]

By application of two of the *three basic rules for differentiation* we can determine the derivative of #a\cdot f+b\cdot g# as follows:

\[\begin{array}{rcl}\left(a\cdot f+b\cdot g\right)' &=&\left(a\cdot f\right)'+\left(b\cdot g\right)'\\ &&\phantom{xxx}\color{blue}{\text{sum rule for }a\cdot f \text{ and }b\cdot g}\\&=&a\cdot f' + b\cdot g'\\&&\phantom{xxx}\color{blue}{\text{product-with-constant rule for }a\cdot f\text{ and } b\cdot g} \end{array}\]

The rule for multiple functions follows by repeatedly applying the rule for sums of two functions.

We can also derive the rule directly: the *difference quotient* of #a\cdot f+b\cdot g# equals

\[\begin{array}{rl}&\dfrac{(a\cdot f+b\cdot g)(c+h) -(a\cdot f+b\cdot g)(c)}{h}\\ &= \dfrac{a\cdot f(c+h)+b\cdot g(c+h)-a\cdot f(c)-b\cdot g(c)}{h}\\ & = a\cdot\dfrac{f(c+h)-f(c)}{h} + b\cdot\dfrac{g(c+h)-g(c)}{h} \end{array}\]

This implies for the derivative:

\[\begin{array}{rcl}(a\cdot f+b\cdot g)'(c)&=&\displaystyle\lim_{h\to 0}\left(a\cdot\dfrac{f(c+h)-f(c)}{h} + b\cdot\dfrac{g(c+h)-g(c)}{h}\right)\\ & =&\displaystyle a\cdot\lim_{h\to 0} \dfrac{f(c+h)-f(c)}{h} + b\cdot\lim_{h\to 0}\dfrac{g(c+h)-g(c)}{h}\\ &=&\displaystyle a\cdot f'(c)+b\cdot g'(c)\end{array}\]

Due to the definition of a sum function, it follows that \[\left(a\cdot f+b\cdot g\right)'(c)= a\cdot f'(c)+b\cdot g'(c)=\left( a\cdot f'+b\cdot g'\right)(c)\tiny.\] This means that the functions #\left(a\cdot f+b\cdot g\right)'# and # a\cdot f'+b\cdot g'# are equal to each other.

By use of the given rule, the derivative of each polynomial can be determined.

After all, by use of the

*extended sum rule*and the

*power rule for differentiation*, we find:

\[ \begin{array}{rcl}\ \frac{\dd}{{\dd}x}

(5\cdot x-3\cdot x^5)&=& -3 \cdot \frac{\dd}{{\dd}x}(x^{5})+5 \cdot \frac{\dd}{{\dd}x}(x) \\

&&\phantom{xx}\color{blue}{\text{extended sum rule}}\\&=& -3\cdot 5 x^{5-1} +5 \cdot 1\\&&\phantom{xx}\color{blue}{\text{power rule}}\\ &=& 5-15\cdot x^4\end{array} \]

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