Calculus

Differentiation of Functions

Definition of the Derivative

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Problems 1-4
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Problems 5-13

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 \(\Delta y\) to the change of the independent variable \(\Delta x\). In the definition of derivative, this ratio is considered in the limit as \(\Delta x \to 0.\) 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
\[
{f’\left( {{x_0}} \right) = \lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} }
= {\lim\limits_{\Delta x \to 0} \frac{{f\left( {{x_0} + \Delta x} \right) – f\left( {{x_0}} \right)}}{{\Delta x}}.}
\] We may denote the derivative in the following ways:
\[f’\left( x \right) = y’\left( x \right) = \frac{{df}}{{dx}} = \frac{{dy}}{{dx}}.\] 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 \({\large\frac{{\Delta y}}{{\Delta x}}\normalsize} = {\large\frac{{f\left( {{x_0} + \Delta x} \right) – f\left( {{x_0}} \right)}}{{\Delta x}}\normalsize}\);
  • 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 \({x_0}\).

In the examples below, 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

Click on problem description to see solution.

 Example 1

Using the definition of derivative, prove that the derivative of a constant is \(0.\)

 Example 2

Calculate the derivative of the function \(y = x.\)

 Example 3

Find the derivative of a linear function \(y = ax + b\) using the definition of derivative.

 Example 4

Using the definition, find the derivative of the simplest quadratic function \(y = {x^2}\).

 Example 5

Determine the derivative of a quadratic function of general form \(y = a{x^2} + bx +c\).

 Example 6

Using the definition of the derivative, find the derivative of the function \(y = \large\frac{1}{x}\normalsize\).

 Example 7

Find the derivative of the function \(y = \sqrt x \).

 Example 8

Calculate the derivative of the cubic function \(y = {x^3}\).

 Example 9

Find the derivative of the sine function \(y = \sin x\).

 Example 10

Find the derivative of the cosine function \(y = \cos x\).

 Example 11

Find an expression for the derivative of the exponential function \(y = {e^x}\) using the definition of derivative.

 Example 12

Find the derivative of the power function \(y = {x^n}\).

 Example 13

Find the derivative of the natural logarithm \(y = \ln x\).

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
\[{y\left( x \right) = C,\;\;\;}\kern-0.3pt {y\left( {x + \Delta x} \right) = C.}\] 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 \(\large\frac{{\Delta y}}{{\Delta x}}\normalsize\) 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.

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 coefficient \(a.\)

Example 4.

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:
\[
{y’\left( x \right) }
= {\lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} }
= {\lim\limits_{\Delta x \to 0} \frac{{\left( {2x + \Delta x} \right)\cancel{\Delta x}}}{{\cancel{\Delta x}}} }
= {\lim\limits_{\Delta x \to 0} \left( {2x + \Delta x} \right) = 2x.}
\]

Page 1
Problems 1-4
Page 2
Problems 5-13