Derivative of a power function

Differentiation: Calculating derivatives and tangent lines

Derivative of a power function

On this page we calculate the derivative of power functions.

Power rule for differentiation

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

$f'(x)=\blue{5}\cdot \green{3}\cdot x^{\green{3}-1}=15\cdot x^2$

Constant
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 $f'(x)=\blue{a}\cdot \green{0} \cdot x^{\green{0}-1}=0$ .

Root

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

$f'(x)=\blue{4}\cdot \green{\frac{1}{2}}\cdot x^{\left(\green{\frac{1}{2}}-1\right)}=2 \cdot x^{-\frac{1}{2}}=\displaystyle\frac{2}{\sqrt{x}}$

Negative power
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

$f'(x)=\blue{3} \cdot (\green{-1}) \cdot x^{\green{-1}-1}=-3\cdot x^{-2}=\displaystyle - \frac{3}{x^2}$

Derivative at a point
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 $f'(2)=15\cdot 2^2=60$ .

Proof 1

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 $\green{n} \geq 1$ .

Proof 2

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 $f(x)=$ $4\cdot x^3$ .

$f'(x)=$ $12\cdot x^2$

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 $a = 4$ and $n = 3$ . So $f'(x)=4\cdot 3 \cdot x^{3-1}=12\cdot x^2$ .

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