Differential Calculus Introduction - SS3 Mathematics Lesson Note

In mathematics, differential calculus is a subfield of mathematics that studies the rates at which quantities change. For example, the rate at which the speed of a vehicle is changing falls under the purview of differential calculus or differentiation.

In calculus, we deal with functions (equations) representing the relationship between two real world quantities. One is an independent variable $x$ and the other is a dependent variable $y$. The function representing their relationship is $y=f(x)$.

For any function, its derivative is defined as the rate of change of the function (dependent variable) with respect to the independent variable. For $y=f(x)$, the derivative $y^{\prime }$ ($y$prime) or $f^{\prime }(x)$ ($f$ prime) is defined as $\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$, the rate of change of $y$ with respect to $x$ as the change in $x$ ($\mathrm{\Delta }x$ tends to $0$).

Differentiation is a process where we find the derivative of a function. It defines the ratio of the change in the value of a function to the change in the independent variable.

Given a function $y=f(x)$, where $y$ is the dependent variable, $x$ is the independent variable,

$\mathrm{\Delta }y$ is the change in the dependent variable, $\mathrm{\Delta }x$ is the change in the independent variable,

$$y=f(x)$$

$$y+\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)$$

$$\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-y$$

$$\mathrm{\Delta }y=f(x+\mathrm{\Delta }x)-f(x)$$

$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$$

$$\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{\mathbf{dy}}{\mathbf{dx}}=f^{{}^{\prime }}(x)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$$

This method of solving for derivatives of functions is known as differentiation by first principle.

Example 1 Find by first principle the differential coefficient of $y=x^2$ with respect to $x$.

Solution

$y=x^2$ i.e. $f(x)=x^2$

If $x$ changes by $\mathrm{\Delta }x$ then $f(x+\mathrm{\Delta }x)=(x+\mathrm{\Delta }x{)}^2$

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}$$

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{(x+\mathrm{\Delta }x{)}^2-x^2}{\mathrm{\Delta }x}$$

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{x^2+2x\mathrm{\Delta }x+(\mathrm{\Delta }x{)}^2-x^2}{\mathrm{\Delta }x}$$

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{2x\mathrm{\Delta }x+(\mathrm{\Delta }x{)}^2}{\mathrm{\Delta }x}$$

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{2x\mathrm{\Delta }x}{\mathrm{\Delta }x}+\frac{(\mathrm{\Delta }x{)}^2}{\mathrm{\Delta }x}$$

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}2x+\mathrm{\Delta }x$$

$\mathrm{\Delta }x$ vanishes as $\mathrm{\Delta }x\to 0$

$$\therefore \frac{dy}{dx}=2x$$