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Application of Derivatives

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  • Last Updated : 28 Jan, 2022
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A function can be considered a machine that receives inputs and produces outputs based on certain conditions. If y is a function of x, then it means that the value of y depends on x. Here, x is the independent variable and y is the dependent variable.

Representation of a function: y = f(x).

Differential Coefficient

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

Mathematical Representation 

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

∆ x = change in x, ∆y = change in y, h = change in x, f(x + h) – f(x) = change in y.

Application of Derivatives

Differentiation is defined as the ratio of change in y to change in x such that the change in x tends to zero. Derivatives play a very important role in the world of Mathematics. They have a wide range of applications in engineering, architecture, economics, and several other fields. Derivatives help business analysts to prepare graphs of profit and loss. 

More than half of the Physics mathematical proofs are based on derivatives. Some of the applications of Derivatives are as follows :

  1. To find the equation of tangent and normal to a curve at a given point.
  2. To find the maxima and minima of a particular function in a given range.
  3. To find the approximate value of quantities.
  4. Derivatives can be used as a rate measure. 

To measure the rate of change of a variable with respect to another variable. Example – Speed = Rate of change of distance with time. v = dx/dt.

  • Significance of derivative as a rate of change. 

It is known that dy / dx = lim h->0  (f(x + h) – f(x) ) / h, here f(x + h) – f(x) is the change in y and (f(x + h) – f(x) ) / h is the rate of change of y with respect to x. This shows that with the help of the concept of derivatives we can find the rate of change of any function. Let’s learn this with the help of an example.

Given, y= 16 – x2. Find the rate of change of y at x = 8.

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

dy/dx = -2 × x [Putting  x = 8]

= -16, hence -16 is the answer.

  • Significance of derivative in finding the approximate value. 

Let’s understand this with the help of an example. Suppose one needs to find the approximate value of √0.037. Let’s consider a function f(x) = √x

f'(x) = (1/2) × x(-1/2) On differentiating f(x) with respect to x.

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

To find √0.037, which can be written as √(0.04 – 0.003)

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

f(0.04-0.003) = 0.1925

√0.037 = 0.1925.                                  

Similarly, one can use differentiation to find the approximation values of functions.

  • Significance of derivative in finding the maxima and minima of any function.

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:

y = 2x3 – 21x2 + 36x – 20.

dy/dx = 6x2 – 42x + 36

Equating dy/dx with 0:

6x2 – 42x + 36 = 0

x2 – 7x + 6 = 0

x2 – (6 + 1)x + 6 = 0

x2 – 6x – x + 6 = 0

x = 6, 1.

The critical points are 6 and 1.

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 QRLTP 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)

= ∆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), 

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 MRLN = 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 equation of tangent MRLN: (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 known 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 equation of normal to the curve at R is given by,

(Y – y) = (-dx/dy) × (X – x), -dx/dy = slope of normal.

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.

Sample Problems

Question 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)

= -(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.                                 

Question 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)

= (- 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

Question 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.

Question 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: (Y – y1) = (dy/dx) × (X – x1)

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

On solving:

(Y × y1) – y12 = (2 × a × X) – (2 × a × x1)

(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

Question 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.                                                   

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