Definition of the Derivative
The derivative of a function is one of the basic concepts of mathematics. Together with the integral, derivative occupies a central place in calculus. The process of finding the derivative is called differentiation. The inverse operation for differentiation is called integration.
The derivative of a function at some point characterizes the rate of change of the function at this point. We can estimate the rate of change by calculating the ratio of change of the function Δy to the change of the independent variable Δx. In the definition of derivative, this ratio is considered in the limit as Δx → 0. Let us turn to a more rigorous formulation.
Formal Definition of the Derivative
Let \(f\left( x \right)\) be a function whose domain contains an open interval about some point \({x_0}\). Then the function \(f\left( x \right)\) is said to be differentiable at \({x_0}\), and the derivative of \(f\left( x \right)\) at \({x_0}\) is given by
\[f'\left( {{x_0}} \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{f\left( {{x_0} + \Delta x} \right) - f\left( {{x_0}} \right)}}{{\Delta x}}.\]
Lagrange's notation is to write the derivative of the function \(y = f\left( x \right)\) as \(f^\prime\left( x \right)\) or \(y^\prime\left( x \right).\)
Leibniz's notation is to write the derivative of the function \(y = f\left( x \right)\) as \(\frac{{df}}{{dx}}\) or \(\frac{{dy}}{{dx}}.\)
The steps to find the derivative of a function \(f\left( x \right)\) at the point \({x_0}\) are as follows:
- Form the difference quotient \(\frac{{\Delta y}}{{\Delta x}} = \frac{{f\left( {{x_0} + \Delta x} \right) - f\left( {{x_0}} \right)}}{{\Delta x}}\);
- Simplify the quotient, canceling \(\Delta x\) if possible;
- Find the derivative \(f'\left( {{x_0}} \right)\), applying the limit to the quotient. If this limit exists, then we say that the function \(f\left( x \right)\) is differentiable at \({x_0}\).
Next, we derive the derivatives of the basic elementary functions using the formal definition of derivative. These functions comprise the backbone in the sense that the derivatives of other functions can be derived from them using the basic differentiation rules.
Solved Problems
Example 1.
Using the definition of derivative, prove that the derivative of a constant is \(0.\)
Solution.
In this case, the function \(y\left( x \right)\) is always equal to to a constant \(C.\) Therefore, we can write
\[y\left( x \right) = C,\;\;\;y\left( {x + \Delta x} \right) = C.\]
It is clear that the increment of the function is identically equal to zero:
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = C - C \equiv 0.\]
Substituting this in the limit definition of derivative, we obtain:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{y\left( {x + \Delta x} \right) - y\left( {x} \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{0}{{\Delta x}} = \lim\limits_{\Delta x \to 0} 0 = 0.\]
Example 2.
Calculate the derivative of the function \(y = x.\)
Solution.
Following the above procedure, we form the ratio \(\frac{{\Delta y}}{{\Delta x}}\) and find the limit as \(\Delta x \to 0:\)
\[\require{cancel} y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\left( {x + \Delta x} \right) - x}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\cancel{x} + \Delta x - \cancel{x}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\cancel{\Delta x}}}{{\cancel{\Delta x}}} = \lim\limits_{\Delta x \to 0} 1 = 1. \]
Example 3.
Using the limit definition find the derivative of the function \(f\left( x \right) = 3x + 2.\)
Solution.
Write the increment of the function:
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = \left[ {3\left( {x + \Delta x} \right) + 2} \right] - \left[ {3x + 2} \right] = \cancel{\color{blue}{3x}} + 3\Delta x + \cancel{\color{red}{2}} - \cancel{\color{blue}{3x}} - \cancel{\color{red}{2}} = 3\Delta x.\]
The difference ratio is equal to
\[\frac{{\Delta y}}{{\Delta x}} = \frac{{3\cancel{\Delta x}}}{{\cancel{\Delta x}}} = 3.\]
Then the derivative is given by
\[f^\prime\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} 3 = 3.\]
Example 4.
Find the derivative of a linear function \(y = ax + b\) using the definition of derivative.
Solution.
We write the increment of the function corresponding to a small change in the argument \(\Delta x:\)
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = \left( {a\left( {x + \Delta x} \right) + b} \right) - \left( {ax + b} \right) = \cancel{\color{blue}{ax}} + a\Delta x + \cancel{\color{red}{b}} - \cancel{\color{blue}{ax}} - \cancel{\color{red}{b}} = a\Delta x.\]
Then the derivative is given by
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{a\cancel{\Delta x}}}{{\cancel{\Delta x}}} = \lim\limits_{\Delta x \to 0} a = a.\]
As it can be seen, the derivative of a linear function \(y = ax + b\) is always constant and equal to the slope coefficient \(a.\)
Example 5.
Using the definition, find the derivative of the simplest quadratic function \(y = {x^2}.\)
Solution.
If we change the independent variable \(x\) by an amount \(\Delta x\), the function receives the following increment:
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = {\left( {x + \Delta x} \right)^2} - {x^2}.\]
This expression can be converted to the form
\[\Delta y = {\left( {x + \Delta x} \right)^2} - {x^2} = \cancel{x^2} + 2x\Delta x + {\left( {\Delta x} \right)^2} - \cancel{x^2} = \left( {2x + \Delta x} \right)\Delta x.\]
By calculating the limit, we find the derivative:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\left( {2x + \Delta x} \right)\cancel{\Delta x}}}{{\cancel{\Delta x}}} = \lim\limits_{\Delta x \to 0} \left( {2x + \Delta x} \right) = 2x.\]
Example 6.
Using the definition of the derivative, differentiate the function \(f\left( x \right) = {x^2} + 2x - 2.\)
Solution.
Calculate the increment of the function:
\[\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) = \left[ {{{\left( {x + \Delta x} \right)}^2} + 2\left( {x + \Delta x} \right) - 2} \right] - \left[ {{x^2} + 2x - 2} \right] = \cancel{\color{darkgreen}{x^2}} + 2x\Delta x + {\left( {\Delta x} \right)^2} + \cancel{\color{blue}{2x}} + 2\Delta x - \cancel{\color{red}{2}} - \cancel{\color{darkgreen}{x^2}} - \cancel{\color{blue}{2x}} + \cancel{\color{red}{2}} = \left( {2x + 2} \right)\Delta x + {\left( {\Delta x} \right)^2}.\]
Write the difference ratio:
\[\frac{{\Delta y}}{{\Delta x}} = \frac{{\left( {2x + 2} \right)\Delta x + {{\left( {\Delta x} \right)}^2}}}{{\Delta x}} = \frac{{\left( {2x + 2 + \Delta x} \right)\cancel{\Delta x}}}{{\cancel{\Delta x}}} = 2x + 2 + \Delta x.\]
Hence, the derivative is
\[f^\prime\left( x \right) = \lim \limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \left( {2x + 2 + \Delta x} \right) = 2x + 2.\]
Example 7.
Determine the derivative of a quadratic function of general form \(y = a{x^2} + bx +c.\)
Solution.
We find the derivative of the given function using the definition of derivative. Write the increment of the function \(\Delta y\) when the argument changes by \(\Delta x:\)
\[\Delta y = y\left( {x + \Delta x} \right) - y\left( x \right) = \left[ {a{{\left( {x + \Delta x} \right)}^2} + b\left( {x + \Delta x} \right) + c} \right] - \left[ {a{x^2} + bx + c} \right] = \cancel{\color{blue}{a{x^2}}} + 2ax\Delta x + a{\left( {\Delta x} \right)^2} + \cancel{\color{red}{bx}} + b\Delta x + \cancel{\color{maroon}{c}} - \cancel{\color{blue}{a{x^2}}} - \cancel{\color{red}{bx}} - \cancel{\color{maroon}{c}} = 2ax\Delta x + a{\left( {\Delta x} \right)^2} + b\Delta x = \left( {2ax + b + a\Delta x} \right)\Delta x. \]
Now we form the ratio of the increments and calculate the limit:
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\left( {2ax + b + a\Delta x} \right)\cancel{\Delta x}}}{{\cancel{\Delta x}}} = \lim\limits_{\Delta x \to 0} \left( {2ax + b + a\Delta x} \right) = 2ax + b. \]
Thus, the derivative of a quadratic function in general form is a linear function.
Example 8.
Using the definition of the derivative, find the derivative of the function \(y = \frac{1}{x}.\)
Solution.
By the definition of derivative,
\[y'\left( x \right) = \lim\limits_{\Delta x \to 0} \frac{{y\left( {x + \Delta x} \right) - y\left( x \right)}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\frac{1}{{x + \Delta x}} - \frac{1}{x}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\frac{{x - \left( {x + \Delta x} \right)}}{{\left( {x + \Delta x} \right)x}}}}{{\Delta x}} = \lim\limits_{\Delta x \to 0} \frac{{\frac{{ - \Delta x}}{{\left( {x + \Delta x} \right)x}}}}{{\Delta x}} = - \lim\limits_{\Delta x \to 0} \frac{\cancel{\Delta x}}{{\cancel{\Delta x}\left( {x + \Delta x} \right)x}} = - \lim\limits_{\Delta x \to 0} \frac{1}{{\left( {x + \Delta x} \right)x}} = - \frac{1}{{{x^2}}}.\]
See more problems on Page 2.
