Application of Derivatives

Last Updated : 06 Jun, 2023

Derivatives are the most important tool in calculus, which has various applications throughout mathematics and beyond mathematics as well. Some fields with various Applications of Derivatives include finance, engineering, and physics as well. Now, let’s say we have a function f(t) which defines the production of toys in a factory somewhere where t is the time for which the factory is operable. Now, using derivatives we can find the maximum and minimum production of toys which can help us plan the purchase order for raw goods and can also run advertisements for the sale of toys whenever we have an excess of toys.

In this article, we will learn about the Application of Derivatives in detail which include the rate of change of quantities, increasing and decreasing functions, approximation, maxima and minima, tangent and normals, etc. By the end of this article, you will have a strong grasp of the subject matter and be able to solve some problems related to various applications of derivatives.

Derivatives Definition

Derivative of a variable y with respect to x is defined as the ratio between the change in y and the change in x, depending upon the condition that changes in x should be very small tending towards zero.

dy/dx = lim ∆x⇢0  ∆y / ∆x = lim h⇢0  (f(x + h) – f(x)) / h.

Where,

• ∆x OR h is change in x, and
• ∆y OR f(x + h) – f(x) is change in y.

Application of Derivatives in Maths

Derivatives play a very important role in the world of Mathematics and also they have various applications beyond mathematics as well i.e., in engineering, architecture, economics, and several other fields. Many proofs of physics and engineering involve applications of derivatives as various physical quantities are the rate change of something over something. For example, velocity is the rate change of displacement, acceleration is the rate change of velocity, etc.

Some of the applications of Derivatives in the field of mathematics are as follows :

Rate Change of Quantities

When two quantities are related by some function then a change in one quantity with respect to another quantity is known as a rate change of quantities and it is represented by the derivatives.

For a function y = f(x), the rate of change of y with respect to x is represented by

dy/dx = lim h⇢0[f(x + h) – f(x)]/h

Where

• h is the rate change in the value of x,
• f(x + h) – f(x) is the change in the value of function i.e., y, and
• [f(x + h) – f(x)]/h is the rate of change of y with respect to x.

Example: For y= 16 – x2. Find the rate of change of y at x = 8.

Solution:

The rate of change of y at x = 8 is given by dy/dx at x = 8,

i.e., dy/dx = -2x [Putting  x = 8]

⇒ dy/dx (at x = 8) = -16,

Hence, -16 is the required answer.

Increasing and Decreasing Function

A function is said to be an increasing function iff for function f(x) if we consider two values in its domain x1 and x2 such that x1>x2, then f(x1)>f(x2). In the words of derivatives, we can define the increasing function as the function for which the slope of its graph is positive i.e., for a function f(x), f'(x) > 0, where f'(x) represents the derivative of the given function.

Example: Check whether function f(x) = x2 is increasing or not for x > 0.

Solution:

for f(x) =  x2

⇒ f'(x) = 2x

Now, for x > 0 f'(x) = 2x, is always positive.

Thus f(x) = x2, is an increasing function for x > 0.

A function is said to be a decreasing function iff for function f(x) if we consider two values in its domain x1 and x2 such that x1>x2, then f(x1) < f(x2). In the words of derivatives, we can define the decreasing function as the function for which the slope of its graph is negative i.e., for a function f(x), f'(x) < 0, where f'(x) represents the derivative of the given function.

Example: Check whether function f(x) = x2 is decreasing or not for x< 0.

Solution:

for f(x) =  x2

⇒ f'(x) = 2x

Now, for x < 0 f'(x) = 2x, is always negative.

Thus f(x) = x2, is an decreasing function for x < 0.

Approximation

As derivative is defined as f'(a) = [f(x) – f(a)]/(x-a)

rearranging the above definition, we find the linear approximation formula for any function f(x).

f(x) ≈ f(a) + f'(a)(x – a)

Example: Approximate the value of √0.037 using derivatives.

Solution:

Let’s consider a function f(x) = √x

On differentiating f(x) with respect to x, we get

f'(x) = (1/2) × x(-1/2)

As 0.037 can rewritten as 0.04 – 0.003,

Thus, h = 0.003

Now, f'(x) = (f(x + h) – f(x)) / h, where h is the change in x.

Thus, -0.003 × (1/2) × 0.04(-1/2) = f(00.4  -0.003) – f(0.04)

⇒ f(0.037) ≈ 0.1925

⇒ √0.037 ≈ 0.1925

Thus, approximation of √0.037  is 0.1925.

Other than a linear approximation, the formula for the quadratic approximation is given as follows:

f(x) ≈ f(a) + f'(a)(x – a) + (1/2)f”(a)(x – a)2

Where,

• f(x), f(a), f'(a), and x are defined as in the linear approximation, and
• f”(a) is the second derivative of the function at a.

Monotonicity

Monotonicity refers to the behaviour of a function, specifically how it changes as its input variable changes. A function is said to be monotonically increasing if its output values increase as its input values increase. Similarly, a function is monotonically decreasing if its output values decrease as its input values increase.

More formally, a function f(x) is said to be:

• Monotonically increasing on an interval I if for any x1, x2 ∈ I such that x1 < x2, we have f(x1) ≤ f(x2).
• Monotonically decreasing on an interval I if for any x1, x2 ∈ I such that x1 < x2, we have f(x1) ≥ f(x2).

Maxima and Minima

The tangent to a curve at the point of maxima or minima is a line parallel to the x-axis. The slope of a line parallel to the x-axis is zero. Hence the value of dy/dx at the point of maxima and minima is zero. Now, the steps involved in finding the point of maxima or minima are as follows:

1. Find the derivative of the function.
2. Equate the derivative with zero to get the critical points.
3. Now find the double derivative of the function.
• If the value of the double derivative at a critical point is less than zero then that point is the point of maxima.
• If the value of the double derivative at a critical point is greater than zero then the point is the point of minima.

Example: Find the local maxima and local minima of the function 2x3 – 21x2 + 36x – 20.

Solution:

Let y = 2x3 – 21x2 + 36x – 20.

⇒ dy/dx = 6x2 – 42x + 36

For Critical point, dy/dx = 0,

⇒ 6x2 – 42x + 36 = 0

⇒ x2 – 7x + 6 = 0

⇒ x2 – (6 + 1)x + 6 = 0

⇒ x2 – 6x – x + 6 = 0

⇒ x = 6, 1.

Thus, the critical points are 6 and 1.

Now, d2y/dx2 = 12x – 42

Putting x = 6.

d2y/dx2 = 12 × 6 – 42 = 30 > 0

Hence, 6 is a point of minima.

Minimum value is 2 × 216 – 21 × 36 + 36 × 6 – 20 = -128

Putting x=1.

d2y/dx2 = 12-42 = -30 < 0

Hence, 1 is apoint of maxima.

Maximum value is 2 – 21 + 36 – 20 = -3.

Tangent and Normal

A line that touches a curve at a point but does not pass through it, is called the tangent to the curve at that point. A normal is a line that is perpendicular to a tangent. The equation of a tangent to a curve is shown in the graph below,

Let y = f(x) be a single-valued function and QRTP be the curve of the function. RT is a chord or a straight line. Coordinates of R = (x, y) and coordinates of T = (x + ∆x, y + ∆y). Slope of a line = m = (y2 – y1) / (x2 – x1),

Slope of RT = (y+∆y – y) / (x+∆x – x)

⇒ Slope of RT = ∆y / ∆x ⇢ (1)

Now, the equation of the chord RT, Y – y = (Slope of RT) × (X – x), [x and y are coordinates of R.]

Y – y = (∆y / ∆x) × (X – x) ⇢ (2),

The slope of RT = ∆y / ∆x.

Now, if the point T gradually moves towards R and in time coincides with R then the chord RT transforms itself to tangent MRLN. This happens when ∆x tends to zero. Therefore equation 2 changes to:

Equation of tangent MN = lim ∆x ⇢ 0   (Y – y) = (∆y / ∆x) × (X – x), This can be written as,

(Y – y) = lim∆x ⇢ 0 (∆y / ∆x) ×  (X – x).

According to the definition of derivatives,

dy/dx = lim∆x ⇢ 0 (∆y / ∆x).

Therefore the equation of tangent MN: (Y – y) = dy/dx × (X – x). This is how we can use the concept of differentiation to find the equation of a tangent to a curve. The normal to a curve is perpendicular to the tangent to the curve.

Note: If two lines are parallel to each other, they both have the same slope. If two lines are perpendicular to each other, the multiplication of their slopes is equal to -1.

As we know that a normal curve is perpendicular to the tangent, therefore,

The slope of normal × Slope of tangent = -1.

Let the slope of normal be m. We know that the slope of tangent = dy/dx. Therefore,

m × dy/dx = -1

⇒ m = – dx/dy

Therefore the equation of normal to the curve at R is given by,

(Y – y1) = (-dx/dy) × (X – x1)

Where, -dx/dy is the slope of normal at (x1, y1

Hence, the concept of the derivatives is used in finding the equations of both the tangent and the normal to a curve at a given point.

Real-Life Applications of Derivatives

Derivatives are mathematical tools used to calculate rates of change. They have numerous real-life applications across various fields.

• Physics: derivatives are used to calculate the velocity and acceleration of moving objects.
• Economics and Finance: derivatives are used to model the behaviour of markets, investments, and other financial instruments. The Black-Scholes equation, which values options, is based on the concept of a derivative.
• Engineering: derivatives are used to optimize designs and control systems. They can be used to calculate the rate of change of temperature in a heat exchanger to optimize its design.
• Medicine: derivatives are used to analyze the behaviour of biological systems, such as the concentration of drugs in the bloodstream over time.
• Computer Science: derivatives are used to optimize algorithms and analyze the performance of the software. They can be used to analyze the time complexity of an algorithm to identify potential inefficiencies and areas for improvement.

Sample Problems on Applications of Derivatives

Problem 1: Find the equation of the tangent and the normal to the circle having equation x2 + y2 = a2 at a point (3, 6).

Solution:

Given, Equation of circle = x2 + y2 = a2.

Differentiating the above equation with respect to x,

2 × x + 2 × y dy/dx = 0

⇒ dy/dx = -(2 × x) / (2 × y)

⇒ dy/dx = -(x / y)

Equation of tangent: (Y-y) = (dy/dx) × (X – x)

⇒ (Y – y) = -(x / y) ×  (X – x)

⇒ (Y × y) – y = -(X × x) + x2 [Multiplying left and right side by y]

⇒ (Y × y) + (X × x) = x2 + y2

⇒ (Y × y) + (X × x) = a2

Putting x = 3 and y = 6,

(Y × 6) + (X × 3) = a2, this is the required equation.

Equation of normal: (Y – y) = (-dx/dy) × (X – x)

⇒ (Y – y) = (y / x) × (X – x), -dx/dy = y / x

⇒ (Y × x) – y × x = (X × y) – y × x

⇒ (Y × x) – (X × y) = 0

Putting x = 3 and y = 6,

(Y × 3) – (X × 6) = 0, this is the required equation.

Problem 2: Find the equation of the tangent to the ellipse having equation (x2 / a2) + (y2 / b2) = 1 at a point (x1, y1).

Solution:

Given, Equation of ellipse = (x2 / a2) + (y2/ b2) = 1

Differentiating the  above the equation with respect to x,

(2 × x) / a2 + ((2 × y) / b2 ) × (dy/dx) = 0

⇒ dy/dx = (-(2 × x) / a2) / ((2 × y) / b2)

⇒ dy/dx = (- x × b2) / (y × a2)

Now, dy/dx at (x1, y1) = (-x1 × b2) / (y1 × a2)

Equation of tangent: (Y – y1) = (dy/dx) × (X – x1)

(Y – y1) = ((-x1 × b2) / (y1 × a2)) × (X – x1)

⇒ (Y × y1 × a2) –  (y12 × a2) = (- X × x1 × b2) + (x12 × b2)

Dividing both sides by (a2 × b2),

((Y × y1) / b2) – (y12 / b2) = -(( X × x1) / a2) + (x12 / a2)

⇒ ((X × x1) / a2) + ((Y × y1) / b2) = (x12 / a2) + (y12 / b2)

⇒ ((X × x1) / a2) + ((Y × y1) / b2) = 1, this is the required equation.

⇒ (x12 / a2) + (y12 / b2) = 1

Problem 3: Find the equation of normal to a curve having equation x2+ y2 – 2 × x – 10 × y  + 16 = 0 at point (2, 2).

Solution:

Given, Equation of curve: x2 + y2 – 2 × x – 10 × y + 16 = 0

Differentiating the equation with respect to x,

2 × x + 2 × y – 2 – (10 × dy/dx) = 0

dy/dx = (- (2 × x) – (2 × y) + 2) / -10

Putting x = 2 and y = 2,

dy/dx = 6/10 = 3 / 5

⇒ -dx/dy = -(5/3)

Equation of  normal: (Y – y) = (-dx/dy) × (X – x)

⇒ (Y – 2) = -(5/3) × (X – 2)

⇒ (3 × Y)  – 6 = (- 5 × X) + 10

⇒ (3 × Y) + (5 × X) = 16, this is the required equation.

Problem 4: Find the equation of the tangent to the parabola having equation y2 = 4 × a × x at the point (x1, y1).

Solution:

Given, Equation of parabola: y2 = 4 × a × x

Differentiating the equation with respect to x,

2 × y × dy/dx = 4 × a

⇒ dy/dx = (4 × a) / (2 × y)

⇒ dy/dx = (4 × a) / (2 × y1) at (x1,y1)

Equation of tangent at (x1, y1) is given by: (Y – y1) = (dy/dx) × (X – x1)

(Y – y1) = ((4 × a) / (2 × y1)) × (X – x1)

⇒ Yy1 – y12 = 2aX – 2ax1

⇒ (Y × y1) – (2 × a × X) – (2 × a × x1) = y12 – (4 × a × x1), subtracting 2 × a × x1 from both sides

(Y × y1) = -2 × a × (X + x1) ,this is the required equation, y12– (4 × a × x1) = 0

Problem 5: Find the equation of the tangent to the curve having equation 4 × x2 + 9 × y2 = 72 at point (3, 2).

Solution:

Given, Equation of curve: 4 × x2 + 9 × y2 = 72

Differentiating the equation with respect to x,

8 × x + 18 × y × dy/dx = 0

dy/dx = (-8 × x) / (18 × y)

Putting x = 3 and y = 2,

dy/dx = -24 / 36 = -2 / 3

Equation of tangent: (Y – 2) = (- 2 / 3) × (X – 3)

(3 × Y) – 6 = (- 2 × X) + 6

(3 × Y) + (2 × X) = 12, this is the required equation.

Note: Trick to write the equation of a tangent to a curve at the point (x1,y1)

1. Replace x2 and y2 in the equation of curve by (x × x1) and (y × y1) respectively.
2. Replace x and y by (x + x1) / 2 and (y + y1) / 2 respectively.
3. Replace (x × y) by ((x × y1) + (y × x1)) / 2
4. Constants remain unchanged.

FAQs on Applications of Derivatives

Q1: What are Derivatives?

Derivatives are defined as the ratio of the rate of change of one variable with respect to the other variable i.e., dy/dx shows the change in the variable y with respect to the change in the variable x.

Q2: What is the Derivative of a Function?

The derivative of a function represents the rate of change of the function at any point on the function. It is defined as the slope of the tangent line to the function at that point.

Q3: How to Find the Derivative of a Function?

Find derivatives there are various ways that can help such as chain rule, quotient rule, power rule, etc. Other than that for a detail description of the same you can read the article “How to find derivative?

Q4: What are Application of Derivatives?

Application of derivatives in mathematics include Rate Change of Quantities, Increasing and Decreasing Function, Approximation, Maxima and Minima, Tangents and Normals Lines, etc.

Q5: What is Difference between First Derivative and Second Derivative of a Function?

The difference between the first and second derivative is simple i.e., when you differenciate the function for the first time, result is called first derivative, and when you again differenciate the first derivative the result is called the second derivative.

Q6: What is Difference between Monotonically Increasing and Increasing Function?

The key difference between monotonically increasing and increasing function is that for increasing function for any two input values x1 and x2  such that x1>x2, increaising function always holds f(x1)>f(x2) but for monotonically increasing function for any two input values x1 and x2  such that x1>x2, monotonically increasing fucntions holds equality as well i.e., f(x1) ≥ f(x2).

Q7: How to Find the Equation of Tangent?

Equation of tangent at a point (x, y) for any function y = f(x) is given as follows:

(Y – y) = dy/dx × (X – x)