The derivative of power functions

Differentiation: The derivative of power functions

The derivative of power functions

In practice, we do not want to calculate derivatives repeatedly using the definition, but we use calculation rules to calculate the derivative directly. We first look at the derivative of a power function.

Power rule

\[\dfrac{\dd}{\dd x}(x^\blue{n})=\blue{n}\cdot x^{\blue{n}-1}\]

Example

\[\begin{array}{rcl} \dfrac{\dd}{\dd x}(x^\blue{5})&=&\blue{5}\cdot x^{\blue{5}-1}\\&=&\blue{5}x^4\end{array}\]

Derivative of the root
\[\dfrac{\dd}{\dd x}\sqrt{x}=\dfrac{\dd}{\dd x}x^{\frac{1}{2}}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}
\]

Proof

Using the definition for the derivative, we find for a power function #f(x)=x^n# where #n# is a positive integer:

\[f'(x)=\dfrac{(x+h)^n-x^n}{h} \text{ where } h \to 0\]

We now expand the brackets:

\[\dfrac{x^n+ n\cdot x^{n-1}\cdot h + \ldots n\cdot x\cdot h^{n-1}+h^n-x^n}{h}\]

The #x^n#-terms cancel:

\[\dfrac{n\cdot x^{n-1}\cdot h + \frac{n(n-1)}{2}\cdot x^{n-2}h^2 + \ldots + n\cdot x\cdot h^{n-1}+h^n}{h}\]

We can write this without a fraction:

\[n\cdot x^{n-1}+ \frac{n(n-1)}{2}\cdot x^{n-2}\cdot h + \ldots + n\cdot x\cdot h^{n-2}+h^{n-1}\]

When we let #h# approach zero, only one term remains: the term without #h#. Therefore, the derivative of #f# is:

\[f'(x)=n\cdot x^{n-1}\]

The proof where #n# is not a positive integer is more advanced and will not be treated here.

If we have a power function with a constant in front of it, we can easily factor out the constant.

Power rule with a constant

\[\dfrac{\dd}{\dd x}(\orange{c}\cdot x^\blue{n})=\orange{c}\cdot\blue{n} \cdot x^{\blue{n}-1}\]

Example

\[\begin{array}{rcl} \dfrac{\dd}{\dd x}(\orange{2}\cdot x^\blue{-3})&=&\orange{2}\cdot\blue{-3}\cdot x^{\blue{-3}-1}\\&=&-6x^{-4}\end{array}\]

Derivative of a constant

The derivative of a constant #\orange{c}# equals #0#. We can show this using #1=x^0# and #\orange{c}=\orange{c}\cdot 1#. This goes as follows:

\[\frac{\dd}{\dd x}\orange{c}=\frac{\dd}{\dd x}(\orange{c}\cdot 1)=\frac{\dd}{\dd x}(\orange{c}\cdot x^0)=\orange{c}\cdot 0 \cdot x^{-1} = 0\]

Calculate #\frac{\dd y}{\dd x}# for \(y = {{2}\over{x^3}} \). Simplify your answer as much as possible and write without negative exponents.

#\frac{\dd y}{\dd x}=# #-{{6}\over{x^4}}#

\[\begin{array}{rcl}
\displaystyle \dfrac{\dd y}{\dd x}&=& \displaystyle \dfrac{\dd}{\dd x}\left( {{2}\over{x^3}} \right)\\
&&\phantom{xxx}\blue{y \text{ substituted}}\\

& =&\displaystyle\dfrac{\dd}{\dd x}\left( 2\cdot x^{-3} \right )\\
&&\phantom{xxx}\blue{\text{rewritten to the form }c\cdot x^n}\\
& =& \displaystyle 2 \cdot -3 \cdot x^{-4}\\
&&\phantom{xxx}\blue{\text{power rule with a constant, }\dfrac{\dd}{\dd x}\left (c \cdot x^n\right)=c \cdot n \cdot x^{n-1}}\\
& =&\displaystyle -{{6}\over{x^4}}\\
&&\phantom{xxx}\blue{\text{simplified}}
\end{array}\]

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