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Application of Derivatives Class 12 Notes CBSE Maths Chapter 6 (Free PDF Download)

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Revision Notes for CBSE Class 12 Maths Chapter 6 (Application of Derivatives) - Free PDF Download

NCERT Maths Class 12 Chapter 6 Application of Derivatives is an important chapter. It is advised that students should revise this chapter as many times as possible. To help students do that, the best quality Class 12 Maths Chapter 6 Revision Notes by Vedantu is the answer. You can also choose to download the  Class 12 Maths Notes Chapter 6 Application of Derivatives Notes PDF for offline revision purposes. Let’s start the NCERT Class 12 Maths Chapter 6 Application of Derivatives Notes by understanding what the term exactly means.


Important Topics Covered in This Chapter

The following are the important topics that are covered under the chapter on the 

Application of Derivatives:

  • Finding the equation of a Tangent and Normal to a Curve.

  • Finding Rate of Change of a Quantity.

  • Finding the Approximation Value.

  • Finding Maxima and Minima, and Point of Inflection.

  • Determining Increasing and Decreasing Functions.


CBSE Class 12 Maths Revision Notes 2024-25 - Chapter Wise PDF Notes for Free

In the table below we have provided the PDF links of all the chapters of CBSE Class 12 Maths whereby the students are required to revise the chapters by downloading the PDF. 


Application of Derivatives Related Important Links

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Application of Derivatives Class 12 Notes Maths - Basic Subjective Questions


Section–A (1 Mark Questions)

1. If $V=\frac{4}{3}\pi r^{3}$ , at what rate in cubic units is V increasing, when r = 10 and $\frac{dr}{dt}=0.01$ ?

Ans. Given: r=10 and $\frac{dr}{dt}=0.01$

We know that $V=\frac{4}{3}\pi r^{3}$ 

$\Rightarrow \frac{dV}{dt}=4\pi r^{3}\frac{dr}{dt}$

$\Rightarrow \frac{dV}{dt}=4\pi (10)^{2}\times0.01$

$\Rightarrow \frac{dV}{dt}=4\pi$.


2. Find the slope of tangent to the curve $y=e^{2x}$   at the point (0, 1).

Ans. Equation of curve: $y=e^{2x}$ 

$\therefore \frac{dy}{dx}=2.e^{2x}$

$\therefore \left ( \frac{dy}{dx} \right )_{0,1}=2.e^{2\times0}=2=$ Slope of tangent to the curve.


3. Find the interval on which the function $f(x)=2x^{2}+9x+12$  is decreasing.

Ans. We have, $f(x)=2x^{2}+9x+12$  

f{}'(x)=4x+9

For f(x) to be decreasing, $f{}'(x)\leq 0$ 

$\Rightarrow 4x+9\leq 0\;\Rightarrow x\leq -\frac{9}{4}$

$\Rightarrow x\epsilon \left ( -\infty ,-\frac{9}{4} \right ]$.


4. Let $f:R\rightarrow R$ be defined by $f(x)=2x+cos\;x$ then prove that f(x)  is an increasing function

Ans. We have, $f(x)=2x+cos\;x$  

$\therefore f{}'(x)=2-sin\;x>0,\;\forall x$

hence, f(x) is an increasing function.


5. Prove that, in a sphere the rate of change of surface area is $8\pi$   times the rate of change of radius.

Ans. Let r be the radius and S be the surface area of the sphere at any time t

Then, $S=4\pi r^{2}$  

$\Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}$. 

$\therefore$ The rate of change of surface area is 8π times the rate of change of the radius.


Section–B (2 Marks Questions)

6. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which the area increases, when side is 10 cm.

Ans. Let the side of triangle $\rightarrow x$ Area $\rightarrow A$ .

Given : $\frac{dx}{dt}=2\;cm/s$

To find : \frac{dA}{dt} at x=10cm

We know that, Area of equilateral triangle, $A=\frac{\sqrt{3}}{4}x^{2}$ 

Differentiating w.r.t. ‘t’, we get

$\frac{dA}{dt}=\frac{\sqrt{3}}{4}.2x.\frac{dx}{dt}=\frac{\sqrt{3}}{2}\frac{dx}{dt}$

At x=10, 

$\frac{dA}{dt}=\frac{10\sqrt{3}}{2}.2=10\sqrt{3}cm^{2}/s$


7. If x is real, then what is the minimum value of x2 – 8x + 17. 

Ans.  We have, $f(x)=x^{2}-8x+17$ 

$\therefore f{}'(x)=2x-8$

$f{}'(x)=0,\;\therefore x=4$

Now, $f{}''(x)=2>0,\;\forall x\;\epsilon R$ 

So, x=4  is the point of minima.

Minimum value of f(x)  at x=4.

$f(4)=4^{2}-8\times4+17=1$.


8. If an error of k% is made in measuring the radius of a sphere, then find the percentage error in its volume.

Ans. Let x be the radius of the sphere and y be its volume. Then,

$\dfrac{\Delta x}{x}\times100=k$

Also, $y=\dfrac{4}{3}\pi x^{3}$ 

$\Rightarrow \dfrac{dy}{dx}=4\pi x^{2}$

$\Rightarrow \dfrac{\Delta y}{y}=\dfrac{4\pi x^{2}}{y}dx=\dfrac{4\pi x^{2}}{\dfrac{4}{3}\pi x^{3}}\times\dfrac{kx}{100}$   $\because \Delta y\approx dy$

$\Rightarrow \dfrac{\Delta y}{y}\times100=3k$

Hence, the error in the volume is 3k%


9. Find the approximate value of $(33)^{1/5}$.

Ans. Consider the function $y=f(x)=x^{\fdrac{1}{5}}$ .

Let: x=32, $\Delta x=1$

Also,

$f(x+\Delta x)=f(x)+f{}'(x)(\Delta x)$

$\Rightarrow (33)^{1/5}=(32)^{1/5}+\dfrac{1}{5(32)^{\dfrac{4}{5}}}(1)$

$\Rightarrow (33)^{1/5}=2+\dfrac{1}{5\times16}$

$\Rightarrow (33)^{1/5}=2.0125$


10. Find the equation of normal to the curve y= tan x at (0,0).

Ans. We have y= tan x

$\Rightarrow \dfrac{dy}{dx}=sec^{2}x$

$\Rightarrow \left ( \dfrac{dy}{dx} \right )_{(0,0)}=sec^{2}0=1$ slope of tangent

 $\Rightarrow$ Slope of normal at point (0,0)  is $=\frac{1}{\left ( \dfrac{dy}{dx} \right )}_{(0,0)}=-\frac{1}{sec^{2}0}=-1$


$\therefore$ Equation of normal :

y-0=-1(x-0)

$\Rightarrow x+y=0$.


11. Find the approximate value of f(3.02), where f(x)=3x2+5x+3.

Ans. Let x=3 and $\Delta x=0.02$ .

We have,

\begin{align*} f(x) &= 3x^2 + 5x + 3 \\ f(3) &= 3 \cdot 3^2 + 5 \cdot 3 + 3 = 45 \\ Now, y &= f(x) \\ \Delta y &= Ay \\ \Delta y &= (6x + 5) \cdot \Delta x \\ \Delta y &= (6 \cdot 3 + 5) \cdot 0.02 = 0.46 \\ f(3.02) &= f(x + \Delta x) = y + \Delta y = 45 + 0.46 = 45.46 \end{align*}

Hence, the approximate value of f(3.02) is 45.46.


12. Find the approximate value of (1.999)5

Ans. Let x=2 and \Delta x=-0.001 

Also let, $y=x^5$

$$\Rightarrow \dfrac{d y}{d x}=5 x^4$$

Now, $$\begin{aligned}& \Delta y=\dfrac{d y}{d x} \Delta x=5 x^4 \times \Delta x=5 \times 2^4 \times[-0.001] \\ & =-80 \times 0.001=-0.080 \\ & \therefore(1.999)^5=y+\Delta y=2^5+(-0.080) \\ & =32-0.080=31.920 \end{aligned} $$


13. Prove that the function $f$ given by $f(x)=\log \sin x$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ and strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$.

Ans. We have, $f(x)=\log \sin x$

$$\therefore f^{\prime}(x)=\dfrac{1}{\sin x} \cos x=\cot x$$

In interval $\left(0, \dfrac{\pi}{2}\right), f^{\prime}(x)=\cot x>0$

$\therefore f$ is strictly increasing in $\left(0, \dfrac{\pi}{2}\right)$

In interval $\left(\dfrac{\pi}{2}, \pi\right), f^{\prime}(x)=\cot x<0$

$\therefore f$ is strictly decreasing in $\left(\dfrac{\pi}{2}, \pi\right)$.


PDF Summary - Class 12 Maths Application of Derivatives Notes (Chapter 6)

1. Derivative as Rate of Change

In various fields of applied mathematics, one has the quest to know the rate at which one variable is changing, with respect to another. The rate of change naturally refers to time. But we can have a rate of change with respect to other variables also. 

  • An economist may want to study how the investment changes with respect to variations in interest rates.

  • A physician may want to know how small changes in dosage can affect the body’s response to a drug.

  • A physicist may want to know the rate of change of distance with respect to time.

All questions of the above type can be interpreted and represented using derivatives.

The average rate of change of a function:

The average rate of change of a function $f(x)$ with respect to $x$ over an interval $[a,a + h]$ is defined as $\dfrac{{f(a + h) - f(a)}}{h}$ .

Instantaneous rate of change of a function:

The instantaneous rate of change of a function$f$ with respect to $x$ is defined as

$f(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$, provided the limit exists.


2. Equations Of Tangent & Normal

(I)The value of the derivative at $P\left( {{x_1},{y_1}} \right)$ gives the slope of the tangent to the curve at\[P\]Symbolically,

${f^\prime }\left( {{x_1}} \right) = {\left. {\dfrac{{dy}}{{dx}}} \right|_{({x_1},{y_1})}} = $ Slope of tangent at $P\left( {{x_1},{y_1}} \right) = m$ (say).

(II) Equation of tangent at $\left( {{x_1},{y_1}} \right)$ is $\left( {y - {y_1}} \right) = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} \times \left( {x - {x_1}} \right)$

(III) Equation of normal at $\left( {{x_1},{y_1}} \right)$ is $\left( {y - {y_1}} \right) = {\left( {\dfrac{{\dfrac{{ - 1}}{{dy}}}}{{dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} \times \left( {x - {x_1}} \right)$


Equations of Tangents and Normal


  • The point $P\left( {{x_1},{y_1}} \right)$ will satisfy the equation of the curve and the equation of tangent and normal line.

  • If the tangent at any point $P$ on the curve is parallel to $X$-axis then $\dfrac{{dy}}{{dx}} = 0$ at the point $P$.

  • If the tangent at any point on the curve is parallel to $Y$-axis, then $\dfrac{{dy}}{{dx}} = \infty \,\,{\text{or}}\,\,\dfrac{{dx}}{{dy}} = 0$.

  • If the tangent at any point on the curve is equally inclined to both the axes then $\dfrac{{dy}}{{dx}} =  \pm 1$.

  • If the tangent at any point makes equal intercept on the coordinate axes then $\dfrac{{dy}}{{dx}} =  \pm 1$.

  • Tangent to a curve at the point $P\left( {{x_1},{y_1}} \right)$ can be drawn even though $\dfrac{{dy}}{{dx}}$ at $P$ does not exist.

Example: $x = 0$ is a tangent to $y = {x^{\dfrac{2}{3}}}$ at $(0,0)$.

  • If a curve passing through the origin is given by a rational integral algebraic equation, the equation of the tangent (or tangents) at the origin is obtained by equating to zero the terms of the lowest degree in the equation.

Example: If the equation of a curve be ${x^2} - {y^2} + {x^3} + 3{x^2}y - {y^3} = 0$, the tangents at the origin are given by ${x^2} - {y^2} = 0$ that is, $x + y = 0$ and $x - y = 0$.

(IV) (a) Length of the tangent $(PT) = \dfrac{{{y_1}\sqrt {1 + {{\left[ {{f^\prime }\left( {{x_1}} \right)} \right]}^2}} }}{{{f^\prime }\left( {{x_1}} \right)}}$

(b) Length of Subtangent $(MT) = \dfrac{{{y_1}}}{{{f^ {'}}\left( {{x_1}} \right)}}$

(c) Length of Normal $(PN) = {y_1}\sqrt {1 + {{\left[ {{f^\prime }\left( {{x_1}} \right)} \right]}^2}} $

(d) Length of Subnormal $(MN) = {y_1}{f^\prime }\left( {{x_1}} \right)$

(V) Differential:

The differential of a function is equal to its derivative multiplied by the differential of the independent variable. 

Thus if, $y = \tan x$ then $dy = {\sec ^2}xdx$

In general, $dy = {f^\prime }(x)dx$.

  • $d(c) = 0$ where ' ${\text{c}}$ ' is a constant.

  • $d(u + v - w) = du + dv - dw$

  • $d(uv) = udv + vdu$

  • The relation $dy = {f^\prime }(x)dx$ can be written as $\dfrac{{dy}}{{dx}} = {f^\prime }({\text{x}})$; thus the quotient of the differentials of $y$ and $x$ is equal to the derivative of $y$ with respect to $x$.


3.Tangent From an External Point

Given a point $P(a,b)$ which does not lie on the curve $y = f(x)$, then the equation of possible tangents to the curve $y = f(x)$, passing through $(a,b)$ can be found by solving for the point of contact $Q$.

And equation of tangent is $y - b = \dfrac{{f(\;h) - b}}{{h - a}}(x - a)$


Tangent from an External Point


4. Angle Between The Curves

The angle between two intersecting curves is defined as the acute angle between their tangents or the normals at the point of intersection of two curves.

$\tan \theta  = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}\;{m_2}}}} \right|$

Here, ${m_1}\,\,{\text{and}}\,\,{{\text{m}}_2}$ are the slopes of tangents at the intersection point $\left( {{x_1},{y_1}} \right)$.


Angle between Curves


  • The angle is defined between two curves if the curves are intersecting. This can be ensured by finding their point of intersection or by graphically.

  • If the curves intersect at more than one point then the angle between curves is found out with respect to the point of intersection.

  • Two curves are said to be orthogonal if the angle between them at each point of intersection is the shortest right angle, that is, ${m_1}\;{m_2} =  - 1$.


5. Shortest Distance Between Two Curves

Shortest distance between two non-intersecting differentiable curves is always along their common normal.


6. Errors and Approximations

(a) Errors

Let $y = f(x)$

From definition of derivative, $\mathop {\lim }\limits_{ax \to 0} \dfrac{{\delta y}}{{\delta x}} = \dfrac{{dy}}{{dx}}$ 

$\dfrac{{\delta y}}{{\delta x}} = \dfrac{{dy}}{{dx}}$ approximately

or $\delta {\text{y}} = \left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right).\delta {\text{x}}$ approximately 

Definition:

(i) $\delta x$ is known as absolute error in $x$.

(ii) $\dfrac{{\delta x}}{x}$ is known as relative error in $x$.

(iii) $\dfrac{{\delta x}}{x} \times 100$ is known as the percentage error in $x$.

  • $\delta {\text{x}}$ and $\delta {\text{y}}$ are known as differentials.


7.Definitions

1.A function $f(x)$ is called an Increasing Function at a point $x$ $ = {\text{a}}$ if in a sufficiently small neighborhood around $x = a$. We have

\[f(a + h) > f(a)\]

\[f(a - h) < f(a)\]

Similarly Decreasing Function if

$f(a + h) < f(a)$

$f(a - h) > f(a)$

Above statements hold true irrespective of whether $f$ is non derivable or even 

discontinuous at $x = a.$

2. A differentiable function is called increasing in an interval $(a,b)$ if it is increasing at every point within the interval (but not necessarily at the endpoints). A function decreasing in an interval $(a,b)$ is similarly defined.

3. A function which in a given interval is increasing or decreasing is called "Monotonic" in that interval.

4. Tests for increasing and decreasing of a function at a point :

If the derivative ${f^\prime }(x)$ is positive at a point $x = a$, then the function $f(x)$ at this point is increasing. If it is negative, then the function is decreasing.

  • Even if ${f^\prime }(a)$ is not defined, $f$ can still be increasing or decreasing.


Increasing and decreasing function


  • ${f^\prime }({\text{c}})$ is not defined in both cases.

If ${f^\prime }(a) = 0$, then for $x = a$ the function may be still increasing or it may be decreasing as shown. It has to be identified by a separate rule.

Example:$f(x) = {x^3}$ is increasing at every point. Note that, $\dfrac{{dy}}{{dx}} = 3{x^2}$


Increasing and decreasing function - 2


  • If a function is invertible it has to be either increasing or decreasing.

  • If a function is continuous, the intervals in which it rises and falls may be separated by points at which its derivative fails to exist.

  • If \[f\] is increasing in $[a,b]$ and is continuous then $f(b)$ is the greatest and $f(a)$ is the least value of $f$ in \[\left[ {a,{\text{ }}b} \right].\] Similarly if $f$ is decreasing in \[[a,b]\] then $f(a)$ is the greatest value and $f(b)$ is the least value.


5. (A) Rolle's Theorem:

Let $f(x)$ be a function of $x$ subject to the following conditions:

(i) $f({\text{x}})$ is a continuous function of ${\text{x}}$ in the closed interval of $a \leqslant x \leqslant b$

(ii) ${f^\prime }({\text{x}})$ exists for every point in the open interval $a < x < b$

(iii) $f({\text{a}}) = f(\;{\text{b}})$

Then there exists at least one point $x = c$ such that $a < c < b$ where ${f^\prime }(c) = 0$

(b) LMVT Theorem:

Let $f(x)$ be a function of $x$ subject to the following conditions:

(i) $f(x)$ is a continuous function of $x$ in the closed interval of $a \leqslant x \leqslant b$.

(ii) ${f^\prime }({\text{x}})$ exists for every point in the open interval $a < x < b$

Then there exists at least one point $x = c$ such that $a < c < b$ where ${f^\prime }(c) = \dfrac{{f(\;b) - f(a)}}{{b - a}}$

Geometrically, the slope of the secant line joining the curve at $x = a\,\,{\text{and}}\,\,x = b$ is equal to the slope of the tangent line drawn to the curve at $x = c$.


Note the following : Rolle's theorem is a special case of LMVT since

$f(a) = f(b) \Rightarrow {f^\prime }(c) = \dfrac{{f(b) - f(a)}}{{b - a}} = 0$.

  • Physical Interpretation of LMVT: Now $[f(b) - f(a)]$ is the change in the function $f$ as $x$ changes from \[a\] to $b$ so that $\dfrac{{f(b) - f(a)}}{{b - a}}$ is the average rate of change of the function over the interval \[\left[ {a,{\text{ }}b} \right].\] Also ${f^\prime }(c)$ is the actual rate of change of the function for $x = c$. Thus, the theorem states that the average rate of change of a function over an interval is also the actual rate of change of the function at some point of the interval. In particular, for instance, the average velocity of a particle over an interval of time is equal to the velocity at some instant belonging to the interval.

This interpretation of the theorem justifies the name "Mean Value" for the theorem.

(c) Application of rolles theorem for isolating the real roots of an equation $f(x) = 0$

Suppose $a\,\& \,\;b$ are two real numbers such that ;

(i) $f(x){\text{ }}\& $ its first derivative ${f^\prime }(x)$ are continuous for $a \leqslant x \leqslant b$

(ii) $f(a)$ and $f(b)$have opposite signs

(iii) \[{f^\prime }(x)\]is different from zero for all values of $x$ between $a\,\& \,b$.

Then there is one and only one real root of the equation $f(x) = 0$ between $a\,\& \,b$.


8.How Maxima & Minima are Classified

1.A function $f(x)$ is said to have a local maximum at $x = a$ if $f(a)$ is greater than every other value assumed by $f(x)$ in the immediate neighbourhood of $x = a$

Symbolically

$\left. {\begin{array}{*{20}{l}} {f(\;a) > f(\;a + h)} \\ {f(\;a) > f(\;a - h)}  \end{array}} \right] $ \Rightarrow x = a$ gives maxima for a sufficiently small positive $h$.

Similarly, a function $f(x)$ is said to have a local minimum value at $x = b$ if $f(\;{\text{b}})$ is least than every other value assumed by $f(x)$ in the immediate neighborhood at $x = b$. Symbolically if

$\left. {\begin{array}{*{20}{l}} {f(\;b) < f(\;b + h)} \\ {f(\;b) < f(\;b - h)}  \end{array}} \right] \Rightarrow x = b$ gives minima for a sufficiently small positive $h$.

(image will be uploaded)


  • The local maximum and local minimum values of a function are also known as local/relative maxima or local/relative minima as these are the greatest and least values of the function relative to some neighborhood of the point in question.

  • The term 'extremum' is used both for maxima or minima.

  • A local maximum (local minimum) value of a function may not be the greatest (least) value in a finite interval.

  • A function can have several local maximum and local minimum values and a local minimum value may even be greater than a local maximum value.

  • Maxima and minima of a continuous function occur alternately and between two consecutive maxima, there is a minima and vice versa.

2. A necessary condition for maxima and minima

If $f(x)$ is a maxima or minima at $x = c$ and if ${f^\prime }(c)$ exists then ${f^\prime }(c) = 0$.

  • The set of values of $x$ for which ${f^\prime }(x) = 0$ are often called stationary points. The rate of change of function is zero at a stationary point.

  • In case ${f^\prime }(c)$ does not exist, $f(c)$ may be a maxima or minima and in this case, left-hand and right-hand derivatives are of opposite signs.

  • The greatest (global maxima) and the least (global minima) values of a function $f$ in an interval \[[a,b]\] are $f(a)$ or $f(b)$ or are given by the values of $x$ which are critical points.

  • Critical points are those where :

      (i) $\dfrac{{dy}}{{dx}} = 0$, if it exists; 

      (ii) or it fails to exist

3.Sufficient condition for extreme values First Derivative Test

$\left. {\begin{array}{*{20}{l}} {{f^\prime }(c - h) > 0} \\  {{f^\prime }(c + h) < 0}  \end{array}} \right] \Rightarrow x = c$ is a point of local maxima

Where $h$ is a sufficiently small positive quantity.

Similarly,

$\left. {\begin{array}{*{20}{l}} {{f^\prime }(c - h) < 0} \\  {{f^\prime }(c + h) > 0}  \end{array}} \right] \Rightarrow x = c$ is a point of local minima,

Where $h$ is a sufficiently small positive quantity.

Note : $ - {f^\prime }(c)$ in both the cases may or may not exist. If it exists, then ${f^\prime }(c) = 0$.

  • If ${f^\prime }(x)$ does not change sign that is, it has the same sign in a certain complete neighbourhood of \[c\], then $f(x)$ is either strictly increasing or decreasing throughout this neighbourhood implying that $f(c)$ is not an extreme value of $f$.

4.Use of second order derivative in ascertaining the maxima or minima

(a) $f(c)$ is a minima of the function $f$, if ${f^\prime }(c) = 0\,\,\& \,\,{f^{\prime \prime }}(c) > 0$

(b) $f(c)$ is a maxima of the function $f$, if ${f^\prime }(c) = 0\,\,\& \,\,{f^{\prime \prime }}(c) < 0$

  • If ${f^{\prime \prime }}(c) = 0$ then the test fails. Revert back to the first-order derivative check for ascertaining the maxima or minima.

5. Summary-working rule

  • First: When possible, draw a figure to illustrate the problem & label those parts that are important in the problem. Constants and variables should be clearly distinguished. 

  • Second: Write an equation for the quantity that is to be maximized or minimized. If this quantity is denoted by $'y'$, it must be expressed in terms of a single independent variable $x$. This may require some algebraic manipulations.

  • Third : If $y = f(x)$ is a quantity to be maximum or minimum, find those values of $x$ for which $\dfrac{{dy}}{{dx}} = {f^\prime }(x) = 0$

  • Fourth: Test each value of $x$ for which ${f^\prime }(x) = 0$ to determine whether it provides a maxima or minima or neither.

The usual tests are :

(a) If $\dfrac{{{d^2}y}}{{d{x^2}}}$ is positive when $\dfrac{{dy}}{{dx}} = 0$

$ \Rightarrow y$ is minima.

If $\dfrac{{{d^2}y}}{{d{x^2}}}$ is negative when $\dfrac{{dy}}{{dx}} = 0$

$ \Rightarrow y$ is maxima.

If $\dfrac{{{d^2}y}}{{d{x^2}}} = 0$ when $\dfrac{{dy}}{{dx}} = 0$, the test fails.

(b) If $\left. {\begin{array}{*{20}{l}} {{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{positive }}}&{{\text{ for }}x < {x_0}} \\  {\dfrac{{dy}}{{dx}}{\text{ is}}\,\,\,\,\,{\text{ zero }}}&{{\text{ for }}x = {x_0}} \\  {{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{negative }}}&{{\text{ for }}x > {x_0}}  \end{array}} \right] \Rightarrow $ a maxima occurs at $x = {x_0}$.

But if $\dfrac{{dy}}{{dx}}$ changes sign from negative to zero to positive as $x$ advances through ${x_0}$, there is a minimum. If $\dfrac{{dy}}{{dx}}$does not change sign, neither a maxima nor a minima. Such points are called INFLECTION POINTS.

  • Fifth : If the function $y = f(x)$ is defined for only a limited range of values $a \leqslant x \leqslant b$ then examine $x = a\,\,\& x = b$ for possible extreme values.

  • Sixth: If the derivative fails to exist at some point, examine this point as possible maxima or minima.

(In general, check at all Critical Points).

  • If the sum of two positive numbers $x$ and $y$ is constant than their product is maximum if they are equal, i.e. $x + y = c,x > 0,y > 0$, then $xy = \dfrac{1}{4}\left[ {{{(x + y)}^2} - {{(x - y)}^2}} \right]$

  • If the product of two positive numbers is constant then their sum is least if they are equal.

${(x + y)^2} = {(x - y)^2} + 4xy$

6. Useful formulae of mensuration to remember

Volume of a cuboid $ = lbh$.

Surface area of a cuboid $ = 2(l\;b + bh + hl)$

Volume of a prism \[ = \] area of the base $ \times $ height.

Lateral surface of a prism \[ = \] perimeter of the base $ \times $ height. Total surface of a prism \[ = \] lateral surface $ + \,\,2$ area of the base

(Note that lateral surfaces of a prism are all rectangles).

Volume of a pyramid $ = \dfrac{1}{3}$ area of the base $ \times $ height.

Curved surface of a pyramid \[ = \]$\dfrac{1}{2}$ (perimeter of the base) $ \times $ slant height.

Volume of a cone $ = \dfrac{1}{3}\pi {r^2}\;h$

Curved surface of a cylinder $ = 2\pi rh$.

Total surface of a cylinder $ = 2\pi rh + 2\pi {r^2}$

Volume of a sphere $ = \dfrac{4}{3}\pi {r^3}$

Surface area of a sphere $ = 4\pi {r^2}$

Area of a circular sector $ = \dfrac{1}{2}{r^2}\theta $, where $\theta $ is in radians.

7.Significance of the sign of 2nd order derivative and points of inflection

The sign of the ${2^{{\text{nd }}}}$ order derivative determines the concavity of the curve. Such points such as $C\,\,\& \,\,E$ on the graph where the concavity of the curve changes are called the points of inflection. From the graph we find that if:

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(i) $\dfrac{{{d^2}}}{{d{x^2}}} > 0 \Rightarrow $ Concave upwards

(ii) $\dfrac{{{d^2}}}{{d{x^2}}} < 0 \Rightarrow $ Concave downwards.

At the point of inflection we find that $\dfrac{{{d^2}}}{{d{x^2}}} = 0$ and $\dfrac{{{d^2}}}{{d{x^2}}}$ changes sign.

Inflection points can also occur if $\dfrac{{{d^2}}}{{d{x^2}}}$ fails to exist (but changes its sign). 

For example, consider the graph of the function defined as,

$f(x) = \left[ {\begin{array}{*{20}{c}} {{x^{3/5}}}&{{\text{  for }}x \in ( - \infty ,1)} \\  {2 - {x^2}}&{{\text{for }}x \in (1,\infty )} \end{array}} \right.$

  • The graph below exhibits two critical points one is a point of local maximum 

$(x = c)$ and the other a point of inflection $(x = 0)$. This implies that not every Critical Point is a point of extrema.


Maxima and Minima Function


Application of Derivatives Class 12 Notes – Chapter Overview

What are Applications of Derivatives in Mathematics?

In earlier chapters, students must have learned how to find the derivatives of different functions, including implicit functions, trigonometric functions, and logarithmic functions. There are many applications of the derivatives of those functions. These applications lie in both mathematical concepts and real-life scenarios. Some of those applications are:

  • Decreasing and increasing functions

  • Newton’s method

  • Linear approximation

  • Rate of change of a quantity

  • Maximum and minimum values

  • Normal and tangent to a curve

There are definitely more applications of derivatives. However, there is one important question that we still have to ask ourselves while going through the revision notes Class 12 Chapter 6. And that question is, what exactly are derivatives?

 

In the simplest terms, derivatives are defined as the rate of change of one quantity to another. When it comes to the terms of functions, then this rate of change of function is depicted as dy / dx = f (x) = y’.

 

Usually, the concept of derivatives is used in both large scale and small scale industries. This concept plays an important role when it comes to determining or bringing about the change of temperature or rate of change of sizes and shapes of an object. This last part depends on various conditions.

 

Before we jump to the next section of these revision notes, Class 12 Math Chapter 6, let’s look at the important applications of derivatives in more depth.

 

Rate of Change of a Quantity

The rate of change of a quantity is the most important and general application of derivatives. For example, if one wants to check the rate of change of the volume of a cube for the decreasing side, then one can use the derivative form of dy / dx.

 

In this equation, dy represents the rate of change of volume of a cube. On the other hand, dx represents the change of the sides of the cube.

 

Decreasing and Increasing Functions

Derivatives are used whenever one wishes to find out whether a given function is decreasing or increasing or it remains constant. This can be done with the help of a graph. Also, if f is a function that is continuous in [p, q] and differentiable in the open interval (p, q), then:

  • F is a constant function in [p, q], if f’ (x) = 0 for each x ∈ (p, q)

  • F is a decreasing function at [p, q] if f’ (x) < 0 for each x ∈ (p, q)

  • F is an increasing function at [p, q] if f’ (x) > ) for every x ∈ (p, q)

 

Learning about Tangent and Normal to a Curve

The next thing that we will learn in these Maths Class 12 Chapter 6 Revision Notes is the application of tangent and normal to a curve. But first, let’s look at the basics. A tangent can be defined as the line that touches the curve at a point. This line doesn’t cross the curve. Also, the normal is the line that is perpendicular to the tangent.

We can also write the straight-line equation that passes through a point, which has a slope m, as:

y- y1 = m (x - x1)

From this equation above, we can see that the slope of the tangent to the curve y = f (x) and at the point P (x1, y1), it is given as dy / dx at P (x1, y1) = f’ (x).

Hence, the equation of the tangent to the curve at P (x1, y1) can also be written as:

y - y1 = f’ (x1) (x - x1)

We can also write the equation of normal to the curve as:

y - y1 = [-1 / f’ (x1)] (x - x1)

The same can also be written as:

(y - y1) f’ (x1) + (x - x1) = 0

 

What are Maxima and Minima?

Another important topic that we need to look at during these NCERT Class 12 Revision Notes Maths Chapter 6 Solution is of minima and maxima. According to notes, the lowest point of a graph is known as minima. And the highest point in a graph is known as maxima.

If an individual wants to calculate the lowest and highest point of the curve in a graph or find the turning point, then a derivative function can be used.

  • When x = a and if f (x) ≤ f (a), for every x in the domain, then f (x) has an absolute maximum value. Also, point a is the point of the maximum value of f

  • When x = a and if f (x) ≥ f (a), for each x in some open interval (p, q), then f (x) has a relative minimum value

  • When x =a and if f (x) ≥ f (a), for each x in the domain, then f (x) has an absolute minimum value. Also, point a is the point for the minimum value of f

  • When x = a and if f (x) ≤ f (a). for each x in some open interval (p, q), then f (x) has a relative maximum value

 

Monotonicity 

Monotonicity is another important topic that students should learn while going through NCERT Solutions Chapter 6 Class 12 Maths Revision Notes. As described in the notes, a function can be said to be monotonic if it has either decreasing or increasing in its entire domain. For example, f (x) = ex, f (x) = nx, and f (x) = 2x + 3. On the other hand, functions that are decreasing or increasing in particular domains are known as non-monotonic. For example, f (x) = x2 and f (x) = sin x.

 

Also, what about the monotonicity of a function at a point? In that case, a function is called monotonically decreasing at x = a, if f (x) satisfies the following conditions:

  • F (x + h) < f (a) for a small position h

  • F’ (x) will be zero if the function is either at its minima or maxima

  • F’ (x) will be the position if the function is increasing

  • F’ (x) will be negative if the function is decreasing

Before going through the Application Of Derivatives Class 12 Maths Revision Notes, students should also know how to find the approximate value or approximation. But what if you have skipped this topic earlier? In that case, readers can learn about this topic now.

 

If one wants to find a very small variation or change in a quantity, then he or she can use derivatives. This would provide the individual with an approximate value of it. The approximate value can also be represented by △ or delta.

 

Let’s assume that change in the value of x, dx = x

In that case, dy / dx = △x = x

This further means that if the change in x, dx is almost equal to x, then dy is also almost equal to y.

 

Point of Inflection

Another important topic that we need to discuss is the point of inflection. In case of a continuous function f (x), if f’ (x0) = 0 or f” (x0) does not exist at locations or points where f’ (x0) exists and if f” (x) changes sign when it is passing through x = x0, then x0 is known as the point of inflection.

There are also two important cases here. And these cases are:

  • If f” (x) < 0, x ∈ (a, b), then the curve y = f (x) is concave downward

  • If f” (x) > 0, x ∈ (a, b), then the curve y = f (x) is concave upward in the case of (a, b)

For example, if we have to solve the equation f (x) = sin x, then we can solve it in the manner that is mentioned below.

F’ (x) = cos x

F” (x) = sin x = 0 x = n π, n ∈ z

 

Fun Facts about Applications of Derivatives

Did you know that derivatives are used in real life to calculate the amount of loss and profit that occurred in a business? This is done by using graphs. You can also check the variations in temperatures by using derivatives. It is also possible to find out the distance or speed covered in terms of kilometers per hour or miles per hour by using derivatives.

 

Apart from all of this, derivatives also play an important role in deriving various equations in the subject of physics. Derivatives also make the study of seismology possible. It can do that by allowing individuals to find the range of magnitudes of any particular earthquake.

Benefits of CBSE Class 12 Revision Notes on Maths Chapter 6 Applications of Derivatives

The following is a list of benefits that you will get by referring to our Revision Notes on Class 12 CBSE Maths Chapter 6 Application of Derivatives:

  • Our revision notes adhere to the rules laid out by the CBSE Board.

  • These Revision Notes for Class 12 CBSE Maths Chapter 6 Application of Derivatives are aligned with the latest CBSE Syllabus and cover all the topics under this chapter.

  • Students studying these revision notes will be able to revise the concepts in a short time, owing to the conciseness of these notes provided.


Conclusion

The Class 12 CBSE Maths Chapter 6 notes on the "Application of Derivatives," available as a free PDF download, are an invaluable resource for both students and educators. These notes dive deep into the practical applications of calculus, revealing how derivatives are employed to analyse and optimise real-world scenarios. With clear explanations and comprehensive problem-solving, they empower students to grasp complex concepts related to rates of change, optimization, and tangent lines. These free PDF notes not only enhance mathematical proficiency but also demonstrate the significance of calculus in various fields like physics, economics, and engineering. In essence, these notes serve as a bridge between theory and practicality, preparing students for advanced studies and real-life problem-solving in mathematics.

FAQs on Application of Derivatives Class 12 Notes CBSE Maths Chapter 6 (Free PDF Download)

1. What is the function of f (x) = x3 - 2 x2 + 2x? It should be noted that x ∈ Q is increasing on Q.

f (x) = x3 - 2 x2 + 2x

If we differentiate both sides, then we will get:

F’ (x) = 3x2 - 4x + 2 > 0 for each value of x

Hence, f is increasing on Q.

2. The tangent to the curve y = x2 - 5x + 5 parallel to the line of 2y = 4x + 1. It also passes through a point. What are the coordinates of the point?

dy / dx = 2x - 5 |x = x1 = 2|

2 x1 = 7

X1 = 7 / 2

Y1 = (49 / 4) - (35 / 2) + 5 = (49 - 70 + 20) / 4 = - 1 / 4

Y + 1 / 4 = 2 (x - 7 / 2)

4y + 1 = 8x - 28

8x - 4y - 29 = 0

X = 1 / 8, y = -7. This would satisfy the equation.

3. The tangent of the curve, y = x ex2, passes through the point (1, e). It also passes through another point. What is that point?

dy / dx = ex2 + + x ex2 .2x

At x = 1, the slope of the tangent m = 3e

The equation of the tangent is:

Y - e = 3e (x - 1)

Y = 3ex - 2e

This means that (4 / 3, 2e) lies on it.

4. Can parents use these notes to assist their child's understanding of the chapter?

Absolutely. Parents can use these notes as a resource to assist their children in understanding the chapter's content and applications, especially if they have a background in mathematics.

6. Can teachers integrate these notes into their classroom teaching of the chapter?

Certainly, teachers can use these notes as a reference to plan lessons and ensure they cover the necessary topics related to the application of derivatives effectively.