On this page we calculate the derivative of power functions.
Let #\blue{a}# and #\green{n}# be real numbers. If \[f(x)=\blue{a}\cdot x^{\green{n}}\] then the derivative of #f# is equal to \[f'(x)=\blue{a}\cdot \green{n}\cdot x^{\green{n}-1}\]
Example
\[\text{For }f(x)=\blue{5}\cdot x^{\green{3}}\] holds \[f'(x)=\blue{5}\cdot \green{3}\cdot x^{\green{3}-1}=15\cdot x^2\]
This rule implies that the derivative of a constant function is equal to zero.
Let #f# be a constant function, so #f(x)=\blue{a}# with real number #\blue{a}#. This function can also be written as #f(x)=\blue{a}\cdot 1 = \blue{a}\cdot x^{\green{0}}#. The derivative of a constant function is thus equal to #f'(x)=\blue{a}\cdot \green{0} \cdot x^{\green{0}-1}=0#.
This rule can also be applied to root functions. A root function can be written as a power function where the exponent #\green{n}# is equal to a fraction.
For example #f(x)=4\cdot\sqrt{x}#. This can be written as \[f(x)=\blue{4}\cdot x^{\green{\frac{1}{2}}}\] Therefore \[f'(x)=\blue{4}\cdot \green{\frac{1}{2}}\cdot x^{\left(\green{\frac{1}{2}}-1\right)}=2 \cdot x^{-\frac{1}{2}}=\displaystyle\frac{2}{\sqrt{x}}\]
This rule can also be applied to negative powers.
For example #f(x)=\displaystyle \frac{3}{x}#. This can be written as \[f(x)=\blue{3}\cdot x^{\green{-1}}\] Therefore \[f'(x)=\blue{3} \cdot (\green{-1}) \cdot x^{\green{-1}-1}=-3\cdot x^{-2}=\displaystyle - \frac{3}{x^2}\]
In the examples we have determined the derivative function, often simply called the derivative. The derivative at a point #x=a# is nothing more than substituting #x=a# into the derivative function.
Take for example #f(x)=\blue{5}\cdot x^{\green{3}}#. We have seen that the derivative function is equal to #f'(x)=15\cdot x^2#. The derivative of #f(x)# at #x=2# is thus #f'(2)=15\cdot 2^2=60#.
Let #\green{n}# be a natural number and #\blue{a}# a real number. We want to show that the statement #\frac{\dd}{\dd x}\left(\blue{a}\cdot x^{\green{n}}\right)=\blue{a}\cdot \green{n}\cdot x^{\green{n}-1}# applies for all natural numbers #\green{n} \geq 1# using
mathematical induction.
Let #\blue{a}# be a real number and #f_\green{n}(x)=\blue{a} \cdot x^\green{n}#. We will first need to verify the base case: \[\lim_{\orange{h} \to 0} \frac{f_\green{1}(x+\orange{h}) - f_\green{1}(x) } {\orange{h}} = \lim_{\orange{h} \to 0} \frac{\blue{a}(x+\orange{h}) - \blue{a} x } {\orange{h}} = \lim_{\orange{h} \to 0} \frac{\blue{a}\orange{h}}{\orange{h}} = \blue{a} \] The limit is a real number and is equal to #\blue{a}#, which means that #f_\green{1}'(x)=\blue{a} = \blue{a} \cdot \green{1} \cdot x^{\green{1}-1}#. The statement thus holds for #\green{n}=1#.
Next we define the induction hypothesis. #f_\green{n}'(x)=\blue{a}\cdot \green{n}\cdot x^{\green{n}-1}#. This means \[\lim_{\orange{h} \to 0} \frac{f_\green{n}(x+\orange{h}) - f_\green{n}(x) } {\orange{h}} =\lim_{\orange{h} \to 0} \frac{\blue{a}(x+\orange{h})^\green{n} - \blue{a} x^\green{n} } {\orange{h}}=\blue{a}\cdot \green{n}\cdot x^{\green{n}-1}\]
Now we can take the induction step, where we want to proof that #f_\green{n+1}'(x)=\blue{a}\cdot (\green{n+1})\cdot x^{\green{n+1}-1}# using the induction hypothesis. We have \[\begin{array}{rcl} \displaystyle \lim_{\orange{h} \to 0} \frac{f_\green{n+1}(x+\orange{h}) - f_\green{n+1}(x) } {\orange{h}} &=& \displaystyle \lim_{\orange{h} \to 0} \frac{\blue{a}(x+\orange{h})^\green{n+1} - \blue{a} x^\green{n+1} }{\orange{h}} \\&&\quad\blue{f_{n+1}\text{ inserted}} \\ &=& \displaystyle \lim_{\orange{h} \to 0} \frac{(x+\orange{h}) \cdot \blue{a}(x+\orange{h})^\green{n} - x \cdot \blue{a} x^\green{n} }{\orange{h}} \\&&\quad\blue{\text{rewritten}} \\ &=& \displaystyle \lim_{\orange{h} \to 0} \left( \left( x \cdot \frac{ \blue{a}(x+\orange{h})^\green{n} - \blue{a} x^\green{n} }{\orange{h}} \right) + \orange{h} \cdot \left(\frac{\blue{a}(x+\orange{h})^\green{n}}{\orange{h}} \right) \right) \\&&\quad\blue{\text{rewritten}} \\ &=& \displaystyle x \cdot \left(\lim_{\orange{h} \to 0} \frac{\blue{a}(x+\orange{h})^\green{n} - \blue{a} x^\green{n} } {\orange{h}} \right) + \lim_{\orange{h} \to 0} \left(\blue{a} (x+\orange{h})^\green{n} \right) \\&&\quad\blue{\text{calculation rules for limits applied and simplified}} \\ &=& x \cdot \blue{a}\cdot \green{n}\cdot x^{\green{n}-1} + \blue{a} \cdot x^\green{n} \\&&\quad\blue{\text{induction hypothesis applied and limit evaluated}} \\ &=& \blue{a} \cdot (\green{n+1}) \cdot x^{\green{n+1} -1} \\&&\quad\blue{\text{simplified}} \end{array}\] The statement therefore also holds for #\green{n+1}# given the induction hypothesis. Through the principles of induction we can thus conclude that the statement works for all #\green{n} \geq 1#.
This second proof also holds for real numbers #\green{n}#, but uses a theory that we are not familiar with yet, namely
implicit differentiation of
logarithmic functions.
Let #\green{n}# and #\blue{a}# be real numbers. We want to prove that #\frac{\dd}{\dd x}\left(\blue{a}\cdot x^{\green{n}}\right)=\blue{a}\cdot \green{n}\cdot x^{\green{n}-1}#. Let #y=\blue{a}\cdot x^{\green{n}}#. Then
\[\begin{array}{rcl}\displaystyle y &=& \blue{a} \cdot x^{\green{n}} \\ &&\quad \blue{\text{given equation}} \\ \ln(y) &=& \ln(\blue{a} \cdot x^{\green{n}}) \\ && \quad \blue{\ln \text{ taken on both sides}} \\ \ln(y) &=& \ln(\blue{a}) + \green{n} \cdot \ln(x) \\ && \quad \blue{\text{calculation rules logarithms}} \\ \displaystyle \frac{1}{y} \frac{\dd y} {\dd x} &=& \displaystyle\frac {\green{n}}{x} \\ && \quad \blue{\text{implicitly differentiated}} \\ \displaystyle\frac{\dd y}{\dd x} &=& \displaystyle y \cdot \frac{\green{n}}{x} \\ && \quad \blue{\text{multiplied both sides with } y } \\ \displaystyle \frac{\dd}{\dd x}\left(\blue{a}\cdot x^{\green{n}}\right) &=&\displaystyle \blue{a} \cdot x^{\green{n}} \cdot \frac{\green{n}}{x} \\ &&\quad \blue{y=a\cdot x^n \text{ substituted}} \\ &=& \blue{a} \cdot \green{n} \cdot x^{\green{n}-1} \\ &&\quad \blue{\text{simplified}}
\end{array}\]
This proofs that #\frac{\dd}{\dd x}\left(\blue{a}\cdot x^{\green{n}}\right)=\blue{a}\cdot \green{n}\cdot x^{\green{n}-1}#.
Determine the derivative of the function #f(x)=# #5\cdot x^4#.
#f'(x)=# #20\cdot x^3#
According to the power rule, the derivative of a function #g(x)=a\cdot x^n# is equal to #g'(x)=a\cdot n\cdot x^{n-1}#. The function #f(x)# is of this form with #a = 5# and #n = 4#. So #f'(x)=5\cdot 4 \cdot x^{4-1}=20\cdot x^3#.
Derivative of a power function
