Differential Calculus is also known as differentiation, which is a very important branch of mathematics known as the mathematics of change. Differential Calculus or Differentiation is used to find out the slope of any line which is the tangent to a given curve at a point on that curve. This statement can be graphically represented as shown below:

## Introduction to Differential Calculus

In the above graph, we have considered a curve y = f (x), that is, y is a function of x. If the change in x is denoted by $\bigtriangleup$ x and the change in y is denoted by $\bigtriangleup$ y, then, we can relate the change in y and x by the following relation:
$\bigtriangleup$ y = (y + $\bigtriangleup$ y) - y = f (x + $\bigtriangleup$ x) - f (x) and $\bigtriangleup$ x = (x + $\bigtriangleup$ x) - x.

Now, if we find the limit of the ratio $\frac{\bigtriangleup y}{\bigtriangleup x}$ as $\bigtriangleup$ x tends to zero, then this ratio is called the first derivative of y with respect to x. This is known as the differential calculus formula, and it is defined by
$f'(x)$ = $\frac{d}{dx}$$f(x) = \lim_{\bigtriangleup x \rightarrow 0 }$$(\frac{\bigtriangleup y}{\bigtriangleup x})$$\lim_{ \bigtriangleup x\rightarrow 0}$$(\frac{f(x+h)-f(x)}{h})$where y is the function of x.

The ratio $\frac{\bigtriangleup y}{\bigtriangleup x}$ also represents the slope of the function f(x) denoted by fâ€™(x), which is the geometrical interpretation of the derivative. Hence, a derivative of a function is the slope of a line tangent to a curve at a given point as shown in the following graph:

Higher order differential coefficients are denoted by $\frac{d^{2}y}{dx^{2}}, \frac{d^{3}y}{dx^{3}}$ etc. which is known as the second derivative, the third derivative and so on.

Derivative of a function can also be defined by considering the definitions of limits, which are defined as follows:

Let f: [a, b] goes to R and a < c < b, then,

Definition 1: The left hand derivative of a function f at x = c, denoted by Lfâ€™(c), is defined as Lfâ€™(c) = $lim_{ x \rightarrow c^-}$$(\frac{f(x)-f(c)}{x-c}), provided that the limit exists. Definition 2: The right hand derivative of a function f at x = c, denoted by Rfâ€™(c), is defined as Rfâ€™(c) = lim_{ x \rightarrow c^+}$$(\frac{f(x)-f(c)}{x-c})$, provided that the limit exists.

Definition 3: If Lfâ€™(c) = Rfâ€™(c), then we say that the function f is derivable at x = c. The common limit is called the derivative of f at x = c and is denoted by fâ€™(c). Thus, fâ€™(c) = $lim_{ x \rightarrow c}$$(\frac{f(x)-f(c)}{x-c}). Definition 4: If Lfâ€™(c) is not equal to Rfâ€™(c), then we say that the function f is not derivable at x = c. Hence, we can conclude that the derivative of any function exist, if and only if, the left hand derivative and the right hand derivative both are equal to each other, and it represents the slope of the curve, written as fâ€™(x). ## Differential Calculus Formulae Many mathematicians have developed various rules for differentiation with which we can find the derivative of any given function. The more commonly used differential formulas are listed below: Derivative of a constant: If a is a constant, then \frac{da}{dx} = 0 or f ‘(a) = 0. This means that the derivative of any number or a constant will always be 0. nth power derivative: \frac{d(x^{n})}{dx} or f‘(xn) = nxn - 1. This means the derivative of a variable which has some constant power is equal to the constant times its variable, provided the power of the variable decreases by 1. For example, consider the derivatives of the following: f‘(x3) = 3x3 - 1 = 3x2 f‘(x-3) = -3x- 3 - 1 = -3x- 4 Multiplication by a constant: If a is a constant, then \frac{d(af(x))}{dx} = a\frac{d(f(x))}{dx}, or f’(af(x)) = af’(f(x)). This means that, if any constant term is multiplied by a function, then it will go outside the derivative of a function. Derivative of a single variable of power one: \frac{d}{dx}$$(x) = 1$. This means that the derivative of a variable having power as one will always be equal to 1. For example, consider f(x) = 2x
f'(x)  = 2

Addition Rule: $\frac{d}{dx}$$[f(x)+g(x)] = \frac{d}{dx}$$f(x)+$$\frac{d}{dx}$$g(x)$, which means the derivative of a sum is the sum of the individual derivatives.

Subtraction Rule: Just like the addition rule, the subtraction rule says, that the derivative of the difference of two functions will be the difference of their individual derivatives, i.e. $\frac{d}{dx}$$[f(x)-g(x)] = \frac{d}{dx}$$f(x)-$$\frac{d}{dx}$$g(x)$

Product Rule: The product rule says that the derivative of the product of two functions will be equal to the second function multiplied with the derivative of the first function plus the first function multiplied with the derivative of the second function, i.e.

$\frac{d}{dx}$$(f(x)g(x)) = f(x)g'(x) + g(x)f'(x) Quotient Rule: The quotient rule says that the derivative of the division of two functions, that is first function divided by the second function, will be equal to the second function multiplied with the derivative of the first function minus the first function multiplied with the derivative of the second function (this will be the numerator), this whole expression divided by the square of the second function, which will be the denominator, i.e. \frac{d}{dx}\frac{f(x)}{g(x)} = \frac{g(x)f'(x)- f(x)g'(x)}{(g(x))^{2}} Derivative of log function: \frac{d}{dx}$$log y$ = $\frac{1}{y} \frac{dy}{dx}$

Derivative of exponent function: $\frac{d}{dx}$$e^{y} = e^{y}$$\frac{dy}{dx}$

Chain Rule: If y = f (g (x)), that is, y is a function of f which is in itself a function of g, then the chain rule is defined as, y’(x) = f’ (g(x)) x g’(x), where g’(x) represents the derivative of function g(x). Chain rule can be summarized as finding the derivative of the outer function first and then, multiplying it with the derivative of the inner function.

The following is the list of some of the trigonometric derivatives formulas:

$\frac{d}{dx}$ sin x = cos x, $\frac{d}{dx}$ cosx = -sinx

$\frac{d}{dx}$ (tan x) = sec2x, $\frac{d}{dx}$ (cot x) = -csc2x

$\frac{d}{dx}$ (sec x) = sec x tan x, $\frac{d}{dx}$ (csc x) = -csc x cot x

## Applications of Differential Calculus

Differential calculus is that branch of mathematics, which is applied in almost every field of science too, whether it is physics or chemistry, or various other research projects.

There are various applications of differentiation as follows:
1. To find out the rate of change of quantities with respect to time or some other variables.
2. To find the equation of the normal and tangent to a curve at a point.
3. To find the maxima and minima of a function.
4. To find out the approximate value of certain quantities, and many more.

We will now explain each of the above mentioned applications in detail as shown below:

Finding the rate of change of quantities with respect to time or some other variables:
A circleâ€™s radius is increasing at the rate of 0.5 cm per second, and if we want to find out its circumferenceâ€™s rate of increase, then by applying the method of derivatives, we can find the rate of change of this quantity as follows:
We know that circumference, $C = 2\pi r$.
Now, the rate of change of this circumference, with respect to time (t) would be:

$\frac{dC}{dt} = \frac{dC}{dr} \times \frac{dr}{dt}$

By applying the chain rule, this implies

$\frac{d}{dr}$$2\pi r$$\frac{dr}{dt}$ = $2\pi $$\frac{dr}{dt} And, it is given in the question that \frac{dr}{dt} = 0.5 cm per second. By substituting this value in the evaluated expression, we would get the rate of increase of the circumference = 2(\pi )(0.5) = \pi cm per second. Finding the equation of the normal and tangent to a curve at a point: Consider a function y = \frac{x^{2}}{2} and we want to find its normal and tangent equation at the point (2, 4). Then, the first step would be to find yâ€™. Hence, \frac{dy}{dx} = \frac{2x}{2} = x. The second step is to find slope of the tangent line, that is \frac{dy}{dx} at x = 2, would be 2. Now, the equation of the tangent is \frac{y-4}{x-2} = 2 and the equation of the slope of the normal is \frac{-1}{2}. Hence, the equation of the normal is \frac{y-4}{x-2} = \frac{-1}{2}. Finding the maxima and minima: We can find the intervals in which the function is increasing or decreasing, as shown below: From the above diagrams, we can conclude that to find the maxima and minima, (together known as extrema) which are very useful to solve optimization problems, the procedure is to first find the first derivative of the function and then equate it (the first derivative) to zero and then solve for values that satisfy the equation. After this, we find its second derivative with the following conclusion: 1. If second derivative > 0, point is a minimum. 2. If second derivative < 0, point is a maximum. ## Differential Calculus Problems Differential calculus problems can easily be solved by understanding the formulas of differentiation, which are very easy to use. The following are some solved problems of differential calculus. ### Solved Examples Question 1: Find the derivatives of the following: 1. 5x^{-3} 2. If f(x) = 3x5 and g(x) = -2x3, then find \frac{d}{dx} (f(x) + g(x)) 3. If f(x) = (x2 + 1) and g(x) = (x3 + 2x), then find \frac{d}{dx} (f(x)g(x)) 4. Differentiate \frac{d}{dx}(\frac{x^{2}+1}{x^{3}+2x}) Solution: By applying the above mentioned list of formulas as required by the question, we will get the following answer. 1. \frac{d}{dx}$$(5x^{-3})$ = 15x-4 by applying the rule of nth derivative of differential calculus.
2. If f(x) = 3x5 and g(x) = -2x3, then $\frac{d}{dx}$ [3x5 + (-2x3)] = 15x4 - 6xby applying the addition rule for differential calculus.
3. If f(x) = (x2 + 1) and g(x) = (x3 + 2x), then $\frac{d}{dx}$ [(x2 + 1) x (x3 + 2x)] = 2x (x3 + 2x) + (3x2 + 2) (x2 + 1), by applying the product rule.
4. $\frac{d}{dx}(\frac{x^{2}+1}{x^{3}+2x})= \frac{2x(x^{3}+2x)-(3x^{2}+2)(x^{2}+1)}{(x^{3}+2x)^{2}}$, by applying the quotient rule of the differential calculus.

Question 2: Find the slope of the tangent line to the curve x2 + 3y2 = 21 at the point (3, -2).
Solution:
The slope of the tangent is given by $\frac{dy}{dx}$.

Hence, 2x + 6y ($\frac{dy}{dx}$) = 0
2 (3) + 6( -2 ) ($\frac{dy}{dx}$) = 0
6 + ( -12 ) ($\frac{dy}{dx}$) = 0
($\frac{dy}{dx}$) = $\frac{1}{2}$
Hence, the slope of the tangent line to the given curve at the point (3, -2) is $\frac{1}{2}$.

Question 3: Find the maximum of the product of two numbers (in which the second number is the square of itself) such that, the sum of the two numbers is 10.
Solution:
Let x and y be two non negative numbers such that, the sum of the two numbers is as follows:10 = x + yy = 10 - x.

Now to find the maximum of their product, we will define the function M as follows:M = xy2 (second number is square of itself).= x(10 - x)2 (By substituting the above value of y)
Differentiating the above function using product and chain rule as required, we will getM' = x (2) (10 - x) (-1) + (1) (10 - x)2    = (10 - x) [-2x + (10 - x)]    = (10 - x) [10 - 3x] = 0

Hence, x = 10 or x = $\frac{10}{3}$ is the answer that will give the maximum product.

## Use of Differential Calculus

Differential calculus is widely used in almost all the areas of science and mathematics, whenever a researcher wants to find the rate of change of its function with respect to time or any other function, or to maximize or minimize the function accordingly. However, differential calculus helps to analyze a graph in detail, as shown below:

As we can see form the above graph that the point of extrema (maxima or minima) is denoted by the pink dots, and we can very easily make out the behavior of the functions, near the points, where the function tends to infinity and where the function is not defined.

With the help of differential calculus, a graphic designer can immediately look and find out that how its design will change rapidly by the changing condition, by just observing the behavior of its graph.

Many Operational research scientists and statisticians widely apply the differential calculus formulae to optimize their problems.

Differential calculus is most importantly, used in physics and other areas of science too as shown below:
If we want to find out the velocity and acceleration of an object by looking at its distance function, then we will compute it as shown below:
Let s (t) be a distance function with respect to time t. Then,
sâ€˜(t) = v(t), which is designated as the instantaneous-velocity function in lieu of time t, and
sâ€˜â€™(t) = vâ€™(t) = a(t), which is denoted as an acceleration-function in lieu of time t.

Let the distance (in meters) which an object travels in the time t seconds is given by:
s(t) = 2t3 + 4t - 5, then s(2), v(2), and a(2) can be calculated as shown below:
s(2) = 2(2)3 + 4(2) - 5 = 16 + 8 - 5 = 19 ft, which gives the distance of an object.
v(t) = sâ€˜(t) = 6t2 + 4, this implies, v(2) = 6(22) +  4 = 28 m per sec, which gives the velocity of the object, and
a(t) = vâ€™(t) = sâ€™â€˜(t) = 12t, this implies, 12 (2) = 24 m per sec square, is the acceleration of the object.

One of the important uses of differential calculus is to find out the types of maxima or minima a function have.

A function can have 4 types of extrema as mentioned below:
• Global maxima
• Local maxima
• Global minima
• Local minima

From the graph, we can conclude that global maxima mean the highest point from all the high points in the graph of a function, whereas global minima is the lowest point of all the low points in the graph. In other words, all the points of inflexion of a curve, above the x axis constitute the local maxima except for the highest point which becomes the global maxima and vice versa. Thus, there can be many local maxima and minima but the global maxima and minima are always unique. This property of differential calculus is widely used in research areas.