Question

How do you find the derivative of \[f\left( x \right) = 2x - 1\]?

Answer 1

Hint: Here, we will use the limit definition formula for the given functions. Then by substituting the limits, we will find the derivative of the function. Differentiation is a method of finding the derivative of the function and finding the rate of change of a function with respect to one variable.

Formula Used:
Limit definition is given by \[f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}}\] .

Complete Step by Step Solution:
We are given a function:
\[f\left( x \right) = 2x - 1\] ………………………………………\[\left( 1 \right)\]
Now, we will find \[f\left( {x + \Delta x} \right)\], so we get
\[ \Rightarrow f\left( {x + \Delta x} \right) = 2\left( {x + \Delta x} \right) - 1\] …………………………………………….\[\left( 2 \right)\]
Now, we will find the derivative of \[2x - 1\] using the limit definition.
By substituting the equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] in the limit definition \[f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f\left( {x + \Delta x} \right) - f\left( x \right)}}{{\Delta x}}\], we get
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{\left( {2\left( {x + \Delta x} \right) - 1} \right) - \left( {2x - 1} \right)}}{{\Delta x}}\]
Multiplying the terms, we get
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{2x + 2\Delta x - 1 - 2x + 1}}{{\Delta x}}\]
Adding and subtracting the like terms, we get
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{2\Delta x}}{{\Delta x}}\]
Now, by simplifying the equation, we get
\[ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{\Delta x \to 0} 2\]
Substituting the limits, we get
\[ \Rightarrow f'\left( x \right) = 2\]

Therefore, the derivative of \[2x - 1\] using the limit definition is \[2\].

Additional information:
We know that the reverse process of differentiation is called antidifferentiation or integration. We should remember some rules in differentiation which include that the derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function. The derivative of a constant is always zero since zero is a constant its derivative is zero.

Note:
We can also solve this question using an alternate method. We will find the derivative by using the differentiation formula.
Now differentiating both sides of the given function \[f\left( x \right) = 2x - 1\] with respect to \[x\], we get
\[ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {2x} \right) - \dfrac{d}{{dx}}\left( 1 \right)\]
Taking the constant out of differentiation, we get
\[ \Rightarrow f'\left( x \right) = 2\dfrac{d}{{dx}}\left( x \right) - \dfrac{d}{{dx}}\left( 1 \right)\]
Now using the differentiation formula \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\] and \[\dfrac{d}{{dx}}\left( k \right) = 0\], we get
\[ \Rightarrow f'\left( x \right) = 2 - 0\]
\[ \Rightarrow f'\left( x \right) = 2\]
Therefore, the derivative of \[2x - 1\] is \[2\].