Applications of Derivatives: Tangents, Maxima-Minima and Explanation

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Derivative comes from the word “derive” meaning to get or obtain something from something else. In Mathematics, Derivative is defined to be an expression that provides us with the rate of change of a function related to an independent variable. One of the most important applications of derivatives is to be found in calculating the rate of change of quantities concerning other quantities.

Various fields of study like Mathematics, Science and Engineering have got applications related to Derivatives.

Read More: Continuity and Differentiability

Key Terms: Intervals, Derivatives, function, Maxima and Minima, First Order Derivative Test, Second Derivative Test.


What is Derivative?

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If some other quantity ‘y’ causes some change in a quantity of certain ‘x’, in view of the fact that an equation of the form y = f(x) gets always satisfied, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}\)

This is also called the Average Rate of Change.

If the rate of change of a function is to be defined at a specific point i.e. a specific value of ‘x’, it is known as the Instantaneous Rate of Change of the function at that point. From the definition of the derivative of a function at a point, we have

\(\frac{dy}{dx}\)|x=x0= Limx→x0 \(\frac{y - y (x_0)}{x - x_0}\)

From this, it is to be concluded that the instantaneous Rate of Change of the function is represented by the derivative of a function. From the rate of change formula, it represents the case when Δx → 0. Thus, the rate of change of ‘y’ with respect to ‘x’ at x = x= (dy/dx)x = x0

Read Also: increasing and decreasing functions


Increasing and Decreasing Function

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In order to understand the real-life processes, one can easily express the form of functions of known variables controlling the processes. One can, then, analyze their behavior as the variable changes to prepare a thorough analysis. We will try to develop more understanding of whether a function increases or decreases by going through what follows:

One can apply this as an extension to the applicability to the Rate of Change quantities we learned a while ago. Let’s take an example of a moving train. If one were to assign a function “speed” to the speed of a moving train and further try to measure the rate of change in speed-related to time as 10 meters / second in a time interval of 1 second, one develops deeper insight into the fact that the function speed acts like an increasing function in the interval of 1 second from its initial point.

The aforementioned analysis helps us understand in figuring out the nature of the speed of the train over different intervals of time irrespective of the values you obtain for the rate of change of speed.

Let y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is called increasing in this interval.
  • Similarly, if for any two points x1 and x2 in the interval x such that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is called decreasing in this interval.

The functions are known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing. Look at the possible shapes of various types of increasing and decreasing functions below:

Increasing and Decreasing Function

Increasing and Decreasing Function

Increasing and Decreasing Function


Monotonic Function

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If a function is differentiable on the interval (a,b) and falls in any one of the four categories explained above i.e. Increasing/strictly Increasing, Decreasing/strictly decreasing, then the function is known as a Monotonic function. Note that if \({\frac{df}{dx} = 0}dxdf=0\), the function is constant on that interval.

The First Derivative Test

The following test finds out the nature of the function, whether it is monotonic or not. The use of the derivative of the function related to the independent variable at a point in the interval is being employed here where the behavior is to be determined. For a function f(x) in the domain (a,b): 

If \(\frac{df}{dx}\)≥ 0 for all x in (a,b), then f(x) is an Increasing Function in (a,b).

Similarly

If, \(\frac{df}{dx}\)≤ 0 for all x in (a,b), then f(x) is an Decreasing Function in (a,b).

For the strictly monotonic nature, strict inequalities must hold i.e.

  • \(\frac{df}{dx}\) > 0 for Strictly Increasing Functions.
  • \(\frac{df}{dx}\)< 0 for Strictly Decreasing Functions.

Proof of the First Derivative Test

We will prove the test for Increasing Functions and leave the part for Decreasing Functions up to you. The test says:

\({{\text{If }}{\frac{df}{dx} \leq 0}}If dxdf≤0 \)for all x in (a,b), then f(x) is an Increasing Function in (a,b). Recount the definition of the derivative of a function:

\(\frac{df}{dx} = \Delta x \)→0 \(\frac{f(x + \Delta x) - f(x)}{\Delta x}\)

If we assume that Δx > 0, then the derivative i.e. \({\frac{df}{dx} /geq 0}\) would be greater than 0 only if the numerator on the R.H.S. of the definition is greater than 0. It would then imply that f(x + Δx) ≥ f(x), which means that f(x) is an increasing function! Similarly, other cases can be proved.

Read More: Differentiation and Integration Formula


Tangents and Normals

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  • Let’s understand this concept with an example of a merry-go-round where the force you experience is towards the center of the merry-go-round but your velocity (the tendency of motion) is in the way towards which your body is pointing.
  • In other words, your velocity at any point is tangential while the force at any point is normal to the circle along which you are moving.
  • This is another branch of derivatives - finding tangents and normals to a given curve. It is a branch of great significance in finding the different maxima and minima of a function, analyzing the directions of velocity and acceleration of a moving object, finding the angles and the shortest distance between two curves, and much more. Let’s jump straight into it!

Tangent

A tangent at a point on the curve is a straight line that touches the curve at that point and whose slope is equal to the gradient/derivative of the curve at that point. From the definition, you can deduce how to find the equation of the tangent to the curve at any point. Given a function y = f(x), the equation of the tangent to this curve at x = x0can be found in the following way:

  • Find out the gradient/derivative of the curve at the point x = x0 : To do this one needs to calculate \(\frac{dy}{dx} \rfloor_{x = x_0}dxdy⌋x=x_0\). Let us call this value m, in analogy to the slope of a straight line.
  • Find the equation of the straight line passing through the point (x0, y(x0)) with slope m. This is quite straightforward and can be found out as:
\(\frac{ y - y_1}{x - x_1} = m\)

Normal

A Normal can be defined as a straight line at a point on the curve that intersects the curve at that point and is perpendicular to the tangent at that point. If one gives its slope by n, and the slope of the tangent at that point or the value of the gradient/derivative at that point is given by m; then we have m×n = -1. Following are the steps to find the normal to a given curve y = f(x) at a point x = x0.

Find out the gradient/derivative of the curve at the point x = x0: This first step is exactly the same as in the method of finding the equation of the tangent to the curve i.e

\(m = \frac{dy}{dx} | x = x_0\)

Find the slope ‘n’ of the normalAs the normal is perpendicular to the tangent, we have:

\(n = \frac{-1}{m}\)

Now, find the equation of the straight line passing through the point (x0, y(x0)) with slope n. The equation is given by:

\(\frac{ y - y_1}{x - x_1} = n\)

Both the tangents and normals to a curve are related. The following diagram explains it better:

Tangents and Normals

Tangents and Normals


Approximations

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  • Approximations can also be simplified, in a layman’s term, rounding off a fractional figure, like saying someone got 98% marks where he scored 97.5 % to be precise.
  • However, the general notion of approximation differs from approximating the values of quantities whose value is unknown or incalculable.
Approximations by Differentials

This method relies on the derivatives of the functions whose values are to be determined at some points.

  • Problem – Given a function y = f(x), determine its value at x = x’.
  • Approach – We will use the definition of the derivative of a function y = f(x) with respect to x. 

\(\frac{d}{dx}\)(f(x)) = change in y with respect to change in x as dx → 0

The change in the value of the function y = f(x) matching to the change dx in x can be obtained if we can obtain the value of x = x’ from a value of x near it, in such a way that the difference in the two values, that is, dx is quite small.

Read More: Differential Equations


General Form

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Notation-wise we will define y(x = x′) = y(x = x0) + Δy; considering that Δy is the change in the value of the function y when the change in x is given by Δx = x′ – x0. After that, we follow the below ;

Find a point x0 near the point x′, at which the value of the function is known

Differentiate the function with respect to x.

\(\frac{dy}{dx} = \frac{d}{dx} (f(x))\)

dy = f’(x)dx

Use the approximations i.e. the value of the change in x i.e. dx = Δx = x′ – x0 and calculate the derivative at x = x0 to get dy, which is approximated as Δy:

Δy = f’(x0) Δx

Δy = f’(x0) (x’ – x0)

This would be the change in the value of the function y as x changes from x0 to x′. Thus, we have

f(x’) = f(x0) + Δy

[f(x’) = f(x0) + f’(x0)(x’ – x0)]

The general form of writing the result obtained above is:

f(x + Δx) = f(x) + f’(x) Δx

which enables one to get the value of the function at a point near x. In connection with this formula, look at the figure below.

General Form

General Form

Read More: logarithmic functions


Maxima and Minima

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This concept finds its application in virtually every domain where one tries to find the minimum and maximum value of a function.

Types of Maxima and Minima

The maxima or minima can also be called an extremum i.e. an extreme value of the function. Let us have a function y = f(x) defined on a known domain of x. Based on the interval of x, on which the function attains an extremum, the extremum can be termed as a ‘local’ or a ‘global’ extremum. We’ll try to understand it in the case of maxima.

  • Local Maxima

A point can be called a Local Maxima when there exists, another point in the domain of a function for which the function value is greater than the local maxima value.

  • Global Maxima

Global Maxima of a function is a point when no other point in the domain of a function exists for which the function value is greater than the global maxima value. 

Look at the graph below to identify the different types of maxima and minima.

Look at the graph below to identify the different types of maxima and minima.

Maxima and Minima

Read Also: tangents and normals


Stationary Points

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A stationary point on a curve is defined as one at which the derivative vanishes i.e. a point (x0, f(x0)) is a stationary point of f(x) if, \([\frac{df}{dx}]_{ x = x_0} = 0\)

Following are the types of stationary points : 

  • Local Maxima
  • Local Minimas
  • Inflection Points

 Let us have a function y = f(x) that attains a Local Maximum at point x = x0. Near the extremum point, the curve will look something like this:

Stationary Points

Stationary Points

The Second Derivative Test

The Second Derivative Test

The Third Case

The Third Case

The Third Case

Also Read:


Things to Remember

  • Derivative is defined as an expression that provides us with the rate of change of a function related to an independent variable.
  • Instantaneous rate of change is the rate of change of a function at a specific point.
  • The different types of functions are increasing function, decreasing function, monotonic function, maxima and minima function and approximations.
  • A Normal can be defined as a straight line at a point on the curve that intersects the curve at that point and is perpendicular to the tangent at that point.

Previous Years Questions

  1. Which the volume increases (in cubic centimeters per minute) when the radius is 5 centimetres is
  2. The equation of the tangent to the conic x2−y2−8x+2y+11=0 at (2,1) is
  3. Angle between y2=x and x2=y at the origin is​
  4. Where [x] denotes the greatest integer =x, is
  5. The function f(x)=x2e−2x,x>0f(x)=x2e−2x,x>0. Then the maximum value of f(x)f(x) is
  6. If there is an error of k% in measuring the edge of a cube, then the percent error in estimating its volume is
  7. The minimum value of xlogx is
  8. For the curve x2+4xy+8y2=64 the tangents are parallel to the x-axis only at the points​
  9. If f(x)=ex(x−2)2 then​
  10. On the interval [0,1], the function x25(1−x)75 takes its maximum value at the point​
  11. The approximate value of f(x)=x3+5x2−7x+9 at x=1.1 is
  12. The maximum value of 4sin2x−12sinx+74sin2x−12sinx+7 is​
  13. The maximum value of f(x)=logxx(x≠0,x≠1) is​
  14. If x2a2+y2b2=1(a>b) and x2−y2=c2 cut at right angles, then​
  15. The point on the parabola y2=64xy2=64x which is nearest to the line 4x+3y+35=04x+3y+35=0 has coordinates​
  16. Let f(x)=x13+x11+x9+x7+x5+x3+x+19. Then f(x)=0 has​
  17. The rate at which the depth of water is increasing, is​

Sample Questions

Ques: What are the common applications of Derivatives? (1 mark)

Ans: As derivatives exemplify slopes, they can be used to find out the maxima and minima of various functions and also, how much a function is going through variation.

Ques: Describe the product rule for the derivative. (1 mark)

Ans: Product rule can be used to find out the derivative of ‘f(x) = x² sin(x)’.

Ques: What’s the Chain Rule? (1 mark)

Ans: The Chain Rule can be applied to find out the derivative of ‘g(x) = sin(x²)’

Ques: Where can the Limits be applied? (1 mark)

Ans: In Maths, Limit is the value that a function approaches as the input. 

Ques: Find all the points of local maxima and local minima of the function f(x) = (x – 1)3 (x + 1)2 (1 mark)
(a) 1, -1, -1/5
(b) 1, -1
(c) 1, -1/5
(d) -1, -1/5

Ans: (a) 1, -1, -1/5

Ques: Find the local minimum value of the function f(x) = sin4x + cos4x, 0 < x < π(1 mark)
(a) 1√2
(b) 12
(c) √32
(d) 0

Ans: (b) 12

Ques: Find the points of local maxima and local minima respectively for the function f(x) = sin 2x – x , where −π2 ≤ x ≤ π(1 mark)
(a) −π6, π6
(b) π3, −π3
(c) −π3, π3
(d) π6, −π6

Ans: (d) π6, −π6

Ques: If y=ax−b(x−1)(x−4) has a turning point P(2, -1), then find the value of a and b respectively. (1 mark)
(a) 1, 2
(b) 2, 1
(c) 0, 1
(d) 1, 0

Ans: (d) 1, 0

Ques: sinp θ cosq θ attains a maximum, when θ = (1 mark)
(a) tan−1pq−−√
(b) tan−1(pq)
(c) tan−1q
(d) tan−1(qp)

Ans: (a) tan−1pq−−√

Ques: Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24x – 18x2(1 mark)
(a) 25
(b) 43
(c) 62
(d) 49

Ans: (d) 49

Ques: If y = x3 + x2 + x + 1, then y (1 mark)
(a) has a local minimum
(b) has a local maximum
(c) neither has a local minimum nor local maximum
(d) None of these

Ans: (c) neither has a local minimum nor local maximum

Ques: Find both the maximum and minimum values respectively of 3x4 – 8x3 + 12x2 – 48x + 1 on the interval [1, 4]. (1 mark)
(a) -63, 257
(b) 257, -40
(c) 257, -63
(d) 63, -257

Ans: (c) 257, -63

Ques: It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value on the interval [0, 2]. Find the value of a. (1 mark)
(a) 100
(b) 120
(c) 140
(d) 160

Ans: (b) 120

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CBSE CLASS XII Related Questions

1.
Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4  \end{bmatrix}\)

      2.
      If (i) A=\(\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix}\),then verify that A'A=I
      (ii) A= \(\begin{bmatrix} \sin\alpha & \cos\alpha\\ -\cos \alpha & \sin\alpha \end{bmatrix}\),then verify that A'A=I

          3.

          Let A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),show that(aI+bA)n=anI+nan-1bA,where I is the identity matrix of order 2 and n∈N

              4.

              If A=\(\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)verify that A3-6A2+9A-4 I=0 and hence find A-1 

                  5.
                  For what values of x,\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)=O?

                      6.
                      Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 &  3\\ 2 & 7\end{bmatrix}\)

                          CBSE CLASS XII Previous Year Papers

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