### Introduction to differentiation: Calculating derivatives

### Derivatives of polynomials and power functions

In the previous sections we have seen how we can calculate the derivative of a function using the definition of the derivative. That is a lot of work. In practice, therefore, rather than using the definition, we calculate the derivative for general standard functions. More complicated functions can be derived by using these known derivatives and calculation rules. Here we discuss the derivative of a polynomial function and a power function.

Three basic rules for differentiating

Let #c# be a real number.

**Constant rule**: The derivative of the constant function #f(x)=c# is #f'(x)=0#.**Product-with-constant rule**: The derivative of the product #c\cdot f(x)# of the constant #c# with a function #f# is #c\cdot f'(x)#.**Sum rule**: If #f# and #g# are functions, then the derivative of the sum function #f(x)+g(x)# equals #f'(x)+g'(x)#.

The constant rule follows from:\[\begin{array}{rcl}f'(x)&=&\frac{\dd}{\dd x} c\\&=&\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\&=&\lim_{h \to 0}\frac{c-c}{h}\\&=&\lim_{h \to 0}0\\&=&0\end{array}\]

The product-with-constant rule follows from:\[\begin{array}{rcl}\frac{\dd}{\dd x}\left(c\cdot f(x)\right)&=&\lim_{h \to 0} \frac{c\cdot f(x+h)-c\cdot f(x)}{h}\\&=&\lim_{h \to 0}c\cdot\frac{ f(x+h)- f(x)}{h}\\&=&c\cdot\lim_{h \to 0}\frac{ f(x+h)- f(x)}{h}\\&=&c\cdot f'(x)\end{array}\]

The sum rule follows from:\[\begin{array}{rcl}\frac{\dd}{\dd x}\left(f(x)+g(x)\right)&=&\lim_{h \to 0} \frac{ f(x+h)+g(x+h)-f(x)-g(x)}{h}\\&=&\lim_{h\to0}\left(\frac{ f(x+h)- f(x)}{h}+\frac{g(x+h)- g(x)}{h}\right)\\&=&f'(x)+g'(x)\end{array}\]

Before we can differentiate polynomials, we have to first learn the derivative of power functions #x^n# where #n# is a natural number. This is a special case of the rule below, which deals with the derivative of all real power functions.

Power rule for differentiation

If #a# is a real number, then the derivative of the function #x^a# equals #ax^{a-1}#. In other words, #\frac{\dd}{{\dd}x}x^a = ax^{a-1}#.

For #a=0#, the function #x^a# is constant, so we already know that its derivative is equal to #0#.

We give a proof of the formula in case #a=n# is a natural number. The *Binomial theorem* gives\[\begin{array}{rcl} \dfrac{(x+h)^n-x^n}{h} &=&\dfrac{x^n+nhx^{n-1} +\frac{n\cdot(n-1)}{2}h^2x^{n-2} + \cdots + nh^{n-1}x+h^n-x^n}{h} \\ &=&\dfrac{nhx^{n-1} +\frac{n\cdot(n-1)}{2}h^2x^{n-2} +\cdots + nh^{n-1}x+h^n}{h}\\ &=&nx^{n-1} + h\cdot \left({\frac{n\cdot(n-1)}{2}x^{n-2} + \cdots + nh^{n-3}x+h^{n-2}}\right)\end{array}\]Taking the limit for #h# to #0#of this expression, we find: \[nx^{n-1} +\lim_{h\to 0}h\cdot\left({\frac{n\cdot(n-1)}{2}x^{n-2}\cdots + nh^{n-3}x+h^{n-2}}\right)=nx^{n-1}\]It follows that the derivative of the function #x^n# is given by #\frac{\dd}{\dd x}x^n=nx^{n-1}#.

By use of the above rule and the sum rule, we can find the derivative of any polynomial:

\[\frac{\dd}{{\dd}x}\left(a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0\right)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1\tiny.\]

Here is a proof by induction on the degree #n# of the polynomial #f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0#:

If #n=0#, then #f(x) =a_0#. Since the derivative of a constant equals #0#, we find \[\frac{\dd}{{\dd}x}\left(f(x)\right)=\frac{\dd}{{\dd}x}\left(a_0\right)=0\tiny,\]from which we conclude that the rule holds for #n=0#.

Now suppose that #n\gt 0#. Write #g(x) = a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots +a_0#. Since the degree of #g(x)# is at most #n-1#, the induction hypothesis applied to #g(x)# gives \[\frac{\dd}{{\dd}x}\left(g(x)\right)=(n-1)a_{n-1}x^{n-2}+(n-2)a_{n-2}x^{n-3}+\cdots+a_1\tiny.\]Using this formula and the *Sum Rule for Differentiation*, we find

\[\begin{array}{rcl} \frac{\dd}{{\dd}x}\left(f(x)\right)&=& \frac{\dd}{{\dd}x}\left(a_nx^{n}\right)+\frac{\dd}{{\dd}x}\left(g(x)\right)\\ &=& \left(na_nx^{n-1}\right)+\left((n-1)a_{n-1}x^{n-2}+(n-2)a_{n-2}x^{n-3}+\cdots+a_1\right)\\ &=& na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1\tiny,\end{array}\]as required.

\[\begin{array}{rcl}f'(x)&=& \dfrac{\dd}{\dd x} x^{99}\\

&=&99 x^{99-1}\\

&&\phantom{xxx}\color{blue}{\text{power rule for differentiation}}\\

&=&99 x^{98}

\end{array}\]

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