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New answer posted

7 months ago

0 Follower 1 View

P
Piyush Vimal

Beginner-Level 5

Students should use our NCERT Applications of Derivatives Solutions to begin their JEE preparation as a resource to build strong foundation.  Our solutions follow the lates CBSE Class 12 syllabus which make them a complete resource for CBSE Boards prepaation, Yet for competitive exam like JEE, Student must go beyond NCERT and refer to advanced-level books like arihant, cengage or coaching material availble to them.

 

New answer posted

7 months ago

0 Follower 1 View

C
Chandra Pruthi

Beginner-Level 5

? Shiksha's NCERT Solutions for Class 12 Applications of Derivatives are designed to serve a holistic mathematical guide for the chapter for all students. That's whay our solutions focus on providing step-by-step textual explanations for each exercise as well as visual representation on 2D graphs along with solutions for better understanding .

We help students with all kinds of problems including those involving increasing/decreasing functions, tangents, normals, and maxima/minima, there is graphical method explanations attached whenever neccessary in image form in the solutions.?

 

New answer posted

7 months ago

0 Follower 8 Views

H
Himanshi Singh

Beginner-Level 5

Class 12 Applications of Derivatives chapter focuses on using derivatives and its concepts to solve real-life problems like rate of change, maxima and minima, increasing/decreasing functions, and finding tangents/normals of the curves at any given co-ordinates. Application of Derivatives is one of the most essential chapter for overall understanding of class 12th calculus unit. 

New answer posted

7 months ago

0 Follower 3 Views

J
Jaya Sinha

Beginner-Level 5

Continuity and Differentiability is one of the most important chapters for CBSE Class 12 Maths board exams. Continuity and Differentiability carries significant weightage of around 6-8 marks in the class 12 board exams. Students can use our Class 12 Math Continuity and Differentiability NCERT Solutions for their benefits, to better preparation and conceptual clarity. students can secure good marks from this chapter by clearing the basics with the help of out NCER Solution prepared by experts in Shiksha.

New answer posted

7 months ago

0 Follower 10 Views

A
Aayush Kumari

Beginner-Level 5

There are several important formulas of the Continuity and Differentiability, Students can cheeck below;

  1. Continuity at a point
    lim?x?a?f(x)=lim?x?a+f(x)=f(a)\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)

  2. Derivative of basic functions:

    • ddx(xn)=nxn?1\frac{d}{dx}(x^n) = nx^{n-1}

    • ddx(sin?x)=cos?x\frac{d}{dx}(\sin x) = \cos x

    • ddx(cos?x)=?sin?x\frac{d}{dx}(\cos x) = -\sin x

    • ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

    • ddx(ln?x)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}

  3. Product Rule:
    ddx[u?v]=u?dvdx+v?dudx\frac{d}{dx}[u \cdot v] = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}

  4. Quotient Rule:
    ddx(uv)=v?dudx?u?dvdxv2\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}

  5. Chain Rule:
    If y=f(g(x))y = f(g(x)), then
    dydx=f?(g(x))?g?(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

New answer posted

7 months ago

0 Follower 1 View

N
nitesh singh

Contributor-Level 10

The concept of continuity is defined for a function  as a curve that is smooth and unbroken at every point in its domain. Mathematically, a function f (x)f (x) is said to be continuous at a point x=ax = a if:

lim? x? af (x)=f (a)\lim_ {x \to a} f (x) = f (a)

What it means:

  • The function is defined at x=ax = a,

  • The limit of the function exists as xx approaches aa,

  • And the limit is equal to the function's value at that point.

Students can check the NCERT Class 12 Maths Chapter 5 Continuity and Differentiability for better and deep understanding of the topic.

New answer posted

7 months ago

0 Follower 11 Views

N
Nishtha Datta

Beginner-Level 5

Yes, Shiksha's NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability are aligned with the latest CBSE 2025 syllabus. Our NCERT Solutions are prepared by subject experts to provide step-by-step explanations, aiding students in their preparation for their board exams. Student can access these solutions and even download the PDF for free on our website Shiksha.

New answer posted

7 months ago

0 Follower 11 Views

S
Satyendra Dhyani

Beginner-Level 5

Students can find well structured NCERT Solutions for Class 12 Continuity and Differentiability in PDF form. We have provided the Solution PDF for all the chapters on our pages, in the Download Free PDF Section. Students can use these NCERT Solutions PDF for their offline study, and as a good and handy resource for practice and revision.

New answer posted

7 months ago

0 Follower 9 Views

E
Esha Garg

Beginner-Level 5

NCERT Class 12 Maths includes Continuity and Differentiability as Chapter 6, which is one of the fundamental concepts in calculus.

  • Continuity is defined for a function in a way that curve of function behaves smoothly without any breaks or jumps at a point or over an interval.
  • Differentiability deal with whether a function has a defined derivative at a point, meaning it has a clear slope or tangent. 

Students can check our NCERT solutions for Continuity and Differentiability for better understanding.

New answer posted

7 months ago

0 Follower 3 Views

P
Piyush Vimal

Beginner-Level 5

Yes, We have updated our NCERT Class 12 Maths Relations and Functions solutions updated as per the latest CBSE 2025 syllabus.We have designed to provide best method of problem-solving which ensures better scoring capabilty of students.These updated solutions help students to do self-preparation, and give confidence while solving board exam questions.

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