NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6

The Central Board of Secondary Education CBSE is one of the largest and most reputable school boards in the country. CBSE is a national board with affiliated schools all over the country. CBSE board has a very objective technique to their approach to education and their curriculum demands a good in-depth understanding of the subject. CBSE serves best to students who want to appear for all different types of competitive exams that take place in our country. The higher secondary curriculum of CBSE is well known all over the country for its high level of difficulty. Many students who transition from a different board in their eleventh standard, have been found to face difficulties in managing the CBSE workload. This might seem extremely challenging, but there are effective ways that can be employed by the candidate to overcome these struggles.

NCERT stands for National Council of Educational Research and Training, and it is a completely resource-based institution founded by the Indian Government. NCERT provides assistance to the Central Government as well as the State Government on academic matters and other matters related to education and the schools that fall under the CBSE board. CBSE board follows all the NCERT prescribed textbooks. The CBSE board focuses on the cognitive development of its students, and therefore, it is a prerequisite for students to follow a pattern of regular revision. Apart from this, extensive hard work, great reference work and a good understanding of all the subjects that they are pursuing are also essential.

Mathematics is a core subject for all the students who have taken Science as their academic stream after they have completed their Class 10 board examinations. The shift from Class 10 to Class 12 has been noted by students and teachers alike to be one of the most difficult transitions a student has to ever go through. Mathematics is already one subject students generally feel great apprehension about, but Class 12 is even more. The CBSE Class 12 Mathematics has a vast and enormous syllabus and finishing the whole curriculum is extremely difficult, apart from finding time to finish revising the syllabus.

NCERT Solutions For Class 12 Maths Chapter 5 Exercise Continuity and Differentiability

One of the major setbacks, according to teachers and students, that students face regarding Mathematics, is primarily the massive syllabus that NCERT has approved for CBSE students, but that is not the only reason. The other reason is the level of difficulty and the student-teacher ratio in most CBSE schools. The increased level of difficulty hinders a smooth progression with the syllabus. Class 12th Math 5.6 is a particularly tricky exercise. Students stumble upon many doubts while solving problems in Exercise 5.6 Class 12th Mathematics. Extramarks has made available the solutions to all the problems in the NCERT Mathematics book. If a student has all the Mathematics problems solved in one place like in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6, they would not have to wait for the teacher’s availability, and they can continue revising by themselves.

These solutions, such as the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6, are provided to the students by leading educators who curate study materials according to specific topics.

List of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 provided by Extramarks follow the best techniques that teachers have unanimously decided to be the most efficient ways to solve the problems of Chapter 5 – Continuity and Differentiability. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 have helped students all over the country and a massive demographic of the students have excelled in their exams. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 not only provides explanations and solutions to every NCERT problem but also explains magnanimously steps that are difficult to follow. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 also provide the list of formulas, concisely presented in every chapter in these solutions. It has been noted that the number of different formulas, cumulative in the entire exercise of NCERT Class 12th Math 5.6, is high. Therefore, the list of all formulas provided in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 has helped students greatly.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 cover all the subsections of Chapter 5. The topics covered in NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 as seen in the NCERT syllabus are as follows –

Algebra of Continuous Functions
Differentiability
Derivatives of Composite Functions
Derivatives of Implicit Functions
Derivatives of Inverse Trigonometric Functions
Exponential and Logarithmic Functions
Logarithmic Differentiation
Derivatives of Functions in Parametric Forms
Second Order Derivative
Mean Value Theorem

Introduction About Derivatives of Functions in Parametric Forms

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 allow the students to learn a different kind of Function called the Parametric Form. In the eleventh standard, students learned about the Introduction to Differentiation and in the next class, students are made aware of other forms of functions.

Parametric Functions are those functions which are described not using general and traditional techniques of defining them explicitly or implicitly but by defining them with a third variable. Learning about a new kind of function can be daunting and implementing complex concepts like Differentiation can seem even more difficult therefore a steady approach is highly advisable for all academics. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6, therefore, focus primarily on the numericals related to the Parametric Functions. Since this is a comparatively new section of the syllabus, the teachers who have compiled and curated the solutions, have ensured that the explanations that are provided are easily understandable and comprehensible by students of all kinds of academic backgrounds.

Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

The solutions to Exercise 5.6 Class 12th Mathematics provided by Extramarks can be accessed through the links provided. The link will direct the students to a PDF which caters completely to the NCERT Class 12 Chapter 5. There are solutions to many numericals in NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 where they could have been solved through multiple methods, but the one provided has been discussed with teachers and educators all over the country. This ensures that the solution provided is the best and that the explanations provided are sufficient for the student to get a good understanding of the chapter, theory, and concepts.

Class 12 Maths Chapter 5 Exercise 5.6 NCERT Solutions: A Brief Introduction

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 have a clear motive of making sure that the student has a good hold over complex concepts like parametric functions. Functions that are formed sometimes have a group of different independent variables and these specific functions are known as the parametric functions. These functions have a very specific use; they are used to represent and define the coordinates of any point for any object that is geometrical like a curve or any other surface. The equations that define these Parametric Functions use Trigonometric Identities and therefore, students need to have a good understanding of Trigonometry is necessary.

These special functions have great applications in the field of science and math because it is through these functions that distinct functions are integrated. If a student wishes to continue their further collegiate and professional degrees in the field of Science, understanding these concepts is crucial and NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6, therefore, provides immense aid to the student.

Why Should You Prefer Using Solutions for Continuity and Differentiability Exercise 5.6?

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 provided by Extramarks ensures that the flow of a student’s academics is not hindered at all. The syllabus of NCERT Mathematics is massive, and the concepts are extremely new therefore it is very understandable and expected that students will have many kinds of doubts and for a teacher to address all individual doubts is extremely difficult. If a student has Extramarks’ NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6, then they automatically get an edge over other candidates. They can directly refer to the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6. More often, it has been observed that while solving an exercise chronologically, students face specific types of doubts. Practising with the help of the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 allows students to assess their preparation levels. Therefore, the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 allow students to save immense time and therefore, expend more time in finishing the syllabus and revising.

Completing Preparing the Chapter Perfectly

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 help provide the scaffold for a self-study session , but doubts can still persist and these doubts can be clarified easily. With the availability of the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6, students can now expedite their revision process and gauge their time management skills.

To prepare the chapter thoroughly, a student has to first solve all the exercises. The doubts that have risen can be answered by referring to the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 have all the formulas concisely. Having a balance between understanding the chapter and its concepts, retaining the formulas and their derivations and solving the sum is absolutely contingent on an optimum understanding of the chapter and in this, the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 help immensely.

Clearing Doubts

Having doubts shows that a student has given much-needed time to these chapters and doubts only arise when they are in the process of problem-solving. Sometimes the doubts can be minor, and sometimes they can be major. Sometimes the doubts arise from mathematical errors, and sometimes they arise because of some incoherence in a student’s understanding of the subject. Understanding what these doubts stand for and what they imply is of utmost importance. Referring to the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 helps many students in contemplating where and why errors persist. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 give students a complete conceptual understanding of the chapter.

Assessing Your Problems – Solving Skills

One of the major difficulties faced by students while following the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 is that they get stuck in the middle of preparation. Mathematics at a higher level gets more complicated because in every step a student has the option of choosing many ways to solve them. Only a few work for a particular sum and one process that works for one problem might not work for a different problem despite being from the same chapter. For higher-level Mathematics, a student has to inculcate the skill to foresee where one approach can lead them and accordingly, plan out the next move and this can be only achieved if a student is in the regular habit of practising numericals. The more a student practises these numericals the more they would be used to the different types of techniques that can be deployed and the more they practice, the better they get at recognising which process to continue with.

Q.1

$\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = a cosθ, y = b cosθ \end{array}$

Ans

$\begin{array}{l} The given equations are x = acosθ, y = bcosθ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} acosθ = - asinθ \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} bcosθ = - bsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- bsinθ}{- asinθ} = \frac{b}{a} \end{array}$

Q.2

\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = sint, y = \cos 2 t \end{array}

Ans

$\begin{array}{l} The given equations are x = sint, y = \cos 2 t \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} sint = cost \\ and \\ \frac{dy}{dt} = \frac{d}{dt} \cos 2 t \\ = - 2 \sin 2 t \\ = - 4 sintcost \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- 4 sintcost}{cost} \\ = - 4 sint \end{array}$

Q.3

\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x= 4 t, y = \frac{4}{t} \end{array}

Ans

$\begin{array}{l} The given equations are x = 4 t, y = \frac{4}{t} \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} 4 t = 4 \\ and \\ \frac{dy}{dt} = \frac{d}{dt} \frac{4}{t} = - \frac{4}{t^{2}} \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{- \frac{4}{t^{2}}}{4} \\ ∴ \frac{dy}{dx} = - \frac{1}{t^{2}} \end{array}$

Q.4

\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = cosθ - \cos 2 θ y = sinθ - \sin 2 θ \end{array}

Ans

$\begin{array}{l} The given equations are x = \cos θ - \cos 2 θ, y = \sin θ - \sin 2 θ \\ Differentiating w .r .t . θ, we get \\ \frac{d x}{d θ} = \frac{d}{d θ} (\cos θ - \cos 2 θ) \\ = - \sin θ + 2 \sin 2 θ \\ a n d \\ \frac{d y}{d θ} = \frac{d}{d θ} (\sin θ - \sin 2 θ) \\ = \cos θ - 2 \cos 2 θ \\ ∴ \frac{d y}{d x} = \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} \\ = \frac{\cos θ - 2 \cos 2 θ}{2 \sin 2 θ - \sin θ} \end{array}$

Q.5

\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = a (θ - sinθ), y = a (1 + cosθ) \end{array}

Ans

$\begin{array}{l} The given equations are x = a (θ - sinθ), y = a (1 + cosθ) \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} a (θ - sinθ) \\ = a (1 - cosθ) \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} a (1 + cosθ) \\ = a (0 - sinθ) \\ = - asinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{- asinθ}{a (1 - cosθ)} \\ = - \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \sin^{2} \frac{θ}{2}} \\ = - \cot \frac{θ}{2} \end{array}$

Q.6

\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}} y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}} \end{array}

Ans

$\begin{array}{l} The given equations are x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}}, y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}} \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} \frac{\sin^{3} t}{\sqrt{\cos 2 t}} \\ = \frac{\sqrt{\cos 2 t} \frac{d}{dt} \sin^{3} t - \sin^{3} t \frac{d}{dt} {(\cos 2 t)}^{\frac{1}{2}}}{{(\sqrt{\cos 2 t})}^{2}} \\ = \frac{\sqrt{\cos 2 t} (3 \sin^{2} t \frac{d}{dt} sint) - \sin^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \frac{d}{dt} \cos 2 t}{\cos 2 t} \\ = \frac{\sqrt{\cos 2 t} (3 \sin^{2} t \times cost) - \sin^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \times - 2 \sin 2 t}{\cos 2 t} \\ = \frac{3 \cos 2 {tsin}^{2} tcost + \sin^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \\ and \\ \frac{dy}{dt} = \frac{d}{dt} \frac{\cos^{3} t}{\sqrt{\cos 2 t}} \\ = \frac{\sqrt{\cos 2 t} \frac{d}{dt} \cos^{3} t - \cos^{3} t \frac{d}{dt} {(\cos 2 t)}^{\frac{1}{2}}}{{(\sqrt{\cos 2 t})}^{2}} \\ = \frac{\sqrt{\cos 2 t} (3 \cos^{2} t \frac{d}{dt} cost) - \cos^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \frac{d}{dt} \cos 2 t}{\cos 2 t} \\ = \frac{\sqrt{\cos 2 t} (3 \cos^{2} t \times - sint) - \cos^{3} t \times \frac{1}{2} {(\cos 2 t)}^{- \frac{1}{2}} \times - 2 \sin 2 t}{\cos 2 t} \\ = \frac{- 3 \cos 2 {tcos}^{2} tsint + \cos^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{\frac{- 3 \cos 2 {tcos}^{2} tsint + \cos^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}}}{\frac{3 \cos 2 {tsin}^{2} tcost + \sin^{3} tsin 2 t}{\cos 2 t \sqrt{\cos 2 t}}} \\ ∴ \frac{dy}{dx} = \frac{- 3 \cos 2 {tcos}^{2} tsint + \cos^{3} t \times 2 sintcost}{3 \cos 2 {tsin}^{2} tcost + \sin^{3} t \times 2 sintcost} [∵ \sin 2 t = 2 sintcost] \\ = \frac{sintcost (- 3 \cos 2 tcost + 2 \cos^{3} t)}{sintcost (3 \cos 2 tsint + 2 \sin^{3} t)} \\ = \frac{- 3 (2 \cos^{2} t - 1) cost + 2 \cos^{3} t}{3 (1 - 2 \sin^{2} t) sint + 2 \sin^{3} t} \\ = \frac{- 6 \cos^{3} t + 3 cost + 2 \cos^{3} t}{3 sint - 6 \sin^{3} t + 2 \sin^{3} t} \\ = \frac{- 4 \cos^{3} t + 3 cost}{3 sint - 4 \sin^{3} t} \\ = - \frac{\cos 3 t}{\sin 3 t} \end{array}$

Q.7

\begin{array}{l} If x and y are connected parametrically by the given equations, without \\ eliminating the parameter, find dydx . \\ x = a (\cos t + \log \tan \frac{t}{2}), y = a \sin t \end{array}

Ans

$\begin{array}{l} The given equations are x = a (cost + logtan \frac{t}{2}), y = asint \\ Differentiating w . r . t . t, we get \\ \frac{dx}{dt} = \frac{d}{dt} a (cost + logtan \frac{t}{2}) \\ = a (\frac{d}{dt} cost + \frac{d}{dt} logtan \frac{t}{2}) \\ = a (- sint + \frac{1}{\tan (\frac{t}{2})} \frac{d}{dt} \tan (\frac{t}{2})) [By chain rule] \\ = a (- sint + \frac{1}{\tan (\frac{t}{2})} . \sec^{2} (\frac{t}{2}) \frac{d}{dt} \frac{t}{2}) \\ = a (- sint + \frac{1}{\tan (\frac{t}{2})} . \sec^{2} (\frac{t}{2}) \times \frac{1}{2}) \\ = a (- sint + \frac{1}{2} \frac{1}{\sin (\frac{t}{2}) \cos (\frac{t}{2})}) \\ = a (- sint + \frac{1}{\sin 2 (\frac{t}{2})}) [∵ \sin 2 θ = 2 sinθcosθ] \\ = a (- sint + \frac{1}{sint}) \\ = a (\frac{1 - \sin^{2} t}{sint}) \\ = a \frac{\cos^{2} t}{sint} \\ and \\ \frac{dy}{dt} = \frac{d}{dt} asint \\ = acost \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{acost}{(a \frac{\cos^{2} t}{sint})} \\ = \frac{sint}{cost} \\ ∴ \frac{dy}{dx} = tant \end{array}$

Q.8 If x and y are connected parametrically by the below equations, without eliminating the parameter, find dy dx.
x = a secθ, y = b tanθ

Ans

$\begin{array}{l} The given equations are x = asecθ, y = b tanθ \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} asecθ \\ = asecθtanθ \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} b tanθ \\ = {bsec}^{2} θ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} \\ = \frac{{bsec}^{2} θ}{asecθtanθ} \\ = \frac{b}{a} . secθ . cotθ \\ = \frac{b}{a} . \frac{1}{cosθ} . \frac{cosθ}{sinθ} \\ = \frac{b}{a} \times \frac{1}{sinθ} \\ = \frac{b}{a} cosecθ \end{array}$

Q.9

\begin{array}{l} If x and y are connected parametrically by the below equations, \\ without eliminating the parameter, find \frac{dy}{dx} . \\ x = a (cosθ + θ sinθ), y = a (sinθ - θ cosθ) \end{array}

Ans

$\begin{array}{l} The given equations are x = a (cosθ + θsinθ), y = a (sinθ - θcos θ) \\ Differentiating w . r . t . θ, we get \\ \frac{dx}{dθ} = \frac{d}{dθ} a (cosθ + θsinθ) \\ = a (- sinθ + θ \frac{d}{dθ} sinθ + sinθ \frac{d}{dθ} θ) \\ = a (- sinθ + θcosθ + sinθ \times 1) \\ = aθcosθ \\ and \\ \frac{dy}{dθ} = \frac{d}{dθ} a (sinθ - θcos θ) \\ = a (cosθ - θ \frac{d}{dθ} cosθ - cosθ \frac{d}{dθ} θ) \\ = a (cosθ + θsinθ - cosθ \times 1) \\ = aθsinθ \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dθ})}{(\frac{dx}{dθ})} = \frac{aθsinθ}{aθcosθ} = tanθ \end{array}$

Q.10

\sqrt{a^{\sin^{- 1} t}}, y = \sqrt{a^{\cos^{- 1} t}}, show that \frac{dy}{dx} = - \frac{y}{x}

Ans

$\begin{array}{l} The given equations are x = \sqrt{a^{\sin^{- 1} t}} = a^{\frac{1}{2} \sin^{- 1} t}, \\ y = \sqrt{a^{\cos^{- 1} t}} = a^{\frac{1}{2} \cos^{- 1} t} \\ Taking logarithm both sides, we get \\ logx = {loga}^{\frac{1}{2} \sin^{- 1} t} \\ = \frac{1}{2} \sin^{- 1} t . loga \\ and \\ logy = {loga}^{\frac{1}{2} \cos^{- 1} t} \\ = \frac{1}{2} \cos^{- 1} t . loga \\ Differentiating w . r . t . t, we get \\ \frac{d}{dt} logx = \frac{d}{dt} \frac{1}{2} \sin^{- 1} t . loga \\ \frac{1}{x} \frac{dx}{dt} = \frac{1}{2} loga \frac{d}{dt} \sin^{- 1} t \\ = \frac{1}{2} loga \times \frac{1}{\sqrt{1 - t^{2}}} \\ \frac{dx}{dt} = x (\frac{loga}{2 \sqrt{1 - t^{2}}}) \\ and \\ \frac{d}{dt} logy = \frac{d}{dt} \frac{1}{2} \cos^{- 1} t . loga \\ \frac{1}{y} \frac{dy}{dt} = \frac{1}{2} loga \frac{d}{dt} \cos^{- 1} t \\ = \frac{1}{2} loga \times \frac{- 1}{\sqrt{1 - t^{2}}} \\ \frac{dy}{dt} = - y (\frac{loga}{2 \sqrt{1 - t^{2}}}) \\ ∴ \frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} \\ = \frac{- y (\frac{loga}{2 \sqrt{1 - t^{2}}})}{x (\frac{loga}{2 \sqrt{1 - t^{2}}})} \\ \Rightarrow \frac{dy}{dx} = - \frac{y}{x} \\ Hence, \frac{dy}{dx} = - \frac{y}{x} \end{array}$

Q.11 Find the second order derivative of the function:
x² + 3x + 2

Ans

$\begin{array}{l} Let y = x^{2} + 3 x + 2 \\ Differentiating both sides w . r . t . x, we get \\ \frac{dy}{dx} = \frac{d}{dx} (x^{2} + 3 x + 2) \\ = 2 x + 3 \\ Differentiating again w . r . t . x, we get \\ ∴ \frac{d^{2} y}{{dx}^{2}} = \frac{d}{dx} (2 x + 3) \\ = 2 \\ Thus, second derivative of given function is 2 . \end{array}$

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6

NCERT Solutions For Class 12 Maths Chapter 5 Exercise Continuity and Differentiability

List of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Introduction About Derivatives of Functions in Parametric Forms

Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Class 12 Maths Chapter 5 Exercise 5.6 NCERT Solutions: A Brief Introduction

Why Should You Prefer Using Solutions for Continuity and Differentiability Exercise 5.6?

Completing Preparing the Chapter Perfectly

Clearing Doubts

Assessing Your Problems – Solving Skills

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6

NCERT Solutions For Class 12 Maths Chapter 5 Exercise Continuity and Differentiability

List of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Introduction About Derivatives of Functions in Parametric Forms

Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Class 12 Maths Chapter 5 Exercise 5.6 NCERT Solutions: A Brief Introduction

Why Should You Prefer Using Solutions for Continuity and Differentiability Exercise 5.6?

Completing Preparing the Chapter Perfectly

Clearing Doubts

Assessing Your Problems – Solving Skills

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FAQs (Frequently Asked Questions)

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