UrbanPro

Learn Unit III: Calculus with Top Tutors

What is your location?

Select Country

search

India

Please enter your locality

Back

Unit III: Calculus

+ Follow 56,799

Top Tutors who teach Unit III: Calculus

1
Sriram B Class 12 Tuition trainer in Vellore Featured
Katpadi, Vellore
Top Tutor
20 yrs of Exp
400per hour
Classes: Class 12 Tuition, Class 9 Tuition and more.

32+ years of teaching

2
Aditi M. Class 12 Tuition trainer in Kolkata Featured
Newtown, Kolkata
10 yrs of Exp
600per hour
Classes: Class 12 Tuition, Class 6 Tuition and more.

I am an experienced, qualified teacher and tutor with over 15 years of experience in teaching maths and chemistry, across different boards including...

3
Ashish Kumar Class 12 Tuition trainer in Bangalore Featured
Mahadevapura, Bangalore
Verified
8 yrs of Exp
1500per hour
Classes: Class 12 Tuition, Quantitative Aptitude and more.

Hi to all, Ashish here i believe in a theory that no matter how much experience you have on paper its of no point unless and until you bring it out...

Do you need help in finding the best teacher matching your requirements?

Post your requirement now
4
Devendra Thakur Class 12 Tuition trainer in Delhi Featured
Vikas Marg, Delhi
20 yrs of Exp
1000per hour
Classes: Class 12 Tuition, Engineering Entrance Coaching and more.

Devendra Thakur - Mathematics Educator With 27 years of teaching experience, I am a seasoned mathematics educator known for a unique approach...

5
Bhagyesh Jain Class 12 Tuition trainer in Hooghly Featured
Uttarpara, Hooghly
Verified
4 yrs of Exp
500per hour
Classes: Class 12 Tuition, Class 11 Tuition and more.

Class 12 Mathematics is just an end of an era and comes with a new era ., in which concepts learned in Class 12 help pupils to decipher Mathematics...

6
Vijay Prajapati Class 12 Tuition trainer in Ranchi Featured
Ranchi Airport, Ranchi
Verified
11 yrs of Exp
400per hour
Classes: Class 12 Tuition, Class 11 Tuition

I have done B Tech in 2011. M. Tech. in 2015 was working as a research professional continuing my PHD from renowned institute CSIR National physical...

7
Ratan Kumar Gupta Class 12 Tuition trainer in Delhi Featured
Uttam Nagar, Delhi
20 yrs of Exp
800per hour
Classes: Class 12 Tuition, Class 9 Tuition and more.

I have 20 yrs experience, taught i many reputed institutes in Delhi

8
Arjun Gautam Class 12 Tuition trainer in Kanpur Featured
Siic IIT Kanpur, Kanpur
5 yrs of Exp
1000per hour
Classes: Class 12 Tuition

I am an Aerospace Engineer, currently pursuing M.Tech from IIT Kanpur. I have 5+ teaching experience in +1 and +2 level Mathematics and Engineering...

9
Subhash K. Class 12 Tuition trainer in Gurgaon Featured
Sector 49 South City II, Gurgaon
Verified
10 yrs of Exp
1000per hour
Classes: Class 12 Tuition, Class 10 Tuition and more.

I cover all the basics concepts of each topic so students understood the topic and enjoy the learning and achieve what they want to achieve, this...

10
Suneel Kumar Class 12 Tuition trainer in Gurgaon Platinum
Baldev Nagar, Gurgaon
Verified
10 yrs of Exp
800per hour
Classes: Class 12 Tuition, Class 9 Tuition and more.

Highlights: •Teacher with 12 years of experience in preparation students for boards •3 month crash course for quick completion and revision of syllabus. •Group...

Guitar Classes in your city

Reviews for top Class 12 Tuition

Average Rating
(4.9)
  • N
    review star review star review star review star review star
    16 Mar, 2013

    Maya attended Class 12 Tuition

    "A very good teacher. "

    V
    review star review star review star review star review star
    19 Mar, 2013

    Swathi attended Class 12 Tuition

    "vijayan sir has immense sincerity towards teaching. He is really good in making concepts..."

    V
    review star review star review star review star review star
    19 Mar, 2013

    Lakshman attended Class 12 Tuition

    "i use to hate phy..when i entered 12th..but after i started my tution with vijayan..."

    V
    review star review star review star review star review star
    20 Mar, 2013

    Hemagowri attended Class 12 Tuition

    "Vijayan Sir is very dedicated and sincere. Teaches the concepts really well and..."

  • A
    review star review star review star review star review star
    29 Mar, 2013

    Student attended Class 12 Tuition

    "Provides complete knowledge for the subject and helps a lot during examination "

    J
    review star review star review star review star review star
    14 Apr, 2013

    Manya attended Class 12 Tuition

    "I learnt a lot and my paper went very well of CBSE 2013.Jagdish explains maths concept..."

    S
    review star review star review star review star review star
    21 Apr, 2013

    Bala attended Class 12 Tuition

    "sir is very good teacher. different short cut methods sir will use.we can learn quikly"

    V
    review star review star review star review star review star
    22 Apr, 2013

    Jayvardhan attended Class 12 Tuition

    "Ya off course his classes are amazing and I had a lot of individual attendence and..."

Get connected

Unit III: Calculus Questions

Ask a Question

Post a Lesson

Answered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability

Sadika

To solve this problem, we will differentiate both sides of the given expression y=(tan⁡−1x)2y=(tan−1x)2 with respect to xx. Let's denote y=(tan⁡−1x)2y=(tan−1x)2. First, let's differentiate yy with respect to xx using the chain rule: dydx=2tan⁡−1x⋅11+x2dxdy=2tan−1x⋅1+x21 Now,... read more

To solve this problem, we will differentiate both sides of the given expression y=(tan⁡−1x)2y=(tan−1x)2 with respect to xx.

Let's denote y=(tan⁡−1x)2y=(tan−1x)2.

First, let's differentiate yy with respect to xx using the chain rule:

dydx=2tan⁡−1x⋅11+x2dxdy=2tan−1x⋅1+x21

Now, let's express dydxdxdy in terms of xx and yy:

dydx=2y⋅11+x2dxdy=2y1+x21

Now, let's differentiate dydxdxdy with respect to xx:

d2ydx2=ddx(2y⋅11+x2)dx2d2y=dxd(2y1+x21)

Using the product rule:

d2ydx2=2⋅11+x2⋅dydx+2y⋅(−2x(1+x2)2)dx2d2y=2⋅1+x21dxdy+2y((1+x2)22x)

Substituting dydx=2y⋅11+x2dxdy=2y1+x21:

d2ydx2=2⋅11+x2⋅2y⋅11+x2+2y⋅(−2x(1+x2)2)dx2d2y=2⋅1+x21⋅2y1+x21+2y((1+x2)22x)

d2ydx2=4y(1+x2)2−4x2y(1+x2)2dx2d2y=(1+x2)24y(1+x2)24x2y

d2ydx2=4y−4x2y(1+x2)2dx2d2y=(1+x2)24y−4x2y

d2ydx2=4(y−x2y)(1+x2)2dx2d2y=(1+x2)24(y−x2y)

Now, let's substitute y=(tan⁡−1x)2y=(tan−1x)2 back into the expression:

d2ydx2=4((tan⁡−1x)2−x2(tan⁡−1x)2)(1+x2)2dx2d2y=(1+x2)24((tan−1x)2−x2(tan−1x)2)

d2ydx2=4(tan⁡−1x)2(1−x2)(1+x2)2dx2d2y=(1+x2)24(tan−1x)2(1−x2)

Now, let's manipulate the expression d2ydx2dx2d2y to match the left side of the given expression (x2+1)2y2+2x(x2+1)y′=2(x2+1)2y2+2x(x2+1)y=2:

(x2+1)2y2+2x(x2+1)y′=(x2+1)2((tan⁡−1x)2)2+2x(x2+1)dydx(x2+1)2y2+2x(x2+1)y=(x2+1)2((tan−1x)2)2+2x(x2+1)dxdy

(x2+1)2y2+2x(x2+1)y′=(x2+1)2(tan⁡−1x)4+2x(x2+1)⋅2y1+x2(x2+1)2y2+2x(x2+1)y=(x2+1)2(tan−1x)4+2x(x2+1)⋅1+x22y

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2(1+x2)2+4x(x2+1)y1+x2(x2+1)2y2+2x(x2+1)y=(1+x2)24(x2+1)2(tan−1x)2+1+x24x(x2+1)y

Now, let's simplify this expression:

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2+4x(x2+1)(tan⁡−1x)2(1+x2)(1+x2)2(x2+1)2y2+2x(x2+1)y=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2+4x(x2+1)(tan⁡−1x)2(1+x2)(1+x2)2(x2+1)2y2+2x(x2+1)y=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)(tan−1x)2(1+x2)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2+4x(x2+1)2(tan⁡−1x)2(1+x2)2(x2+1)2y2+2x(x2+1)y=(1+x2)24(x2+1)2(tan−1x)2+4x(x2+1)2(tan−1x)2

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2(1+x2+x)(1+x2)2(x2+1)2y2+2x(x2+1)y=(1+x2)24(x2+1)2(tan−1x)2(1+x2+x)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)2(tan⁡−1x)2(x2+1)(1+x2)2(x2+1)2y2+2x(x2+1)y=(1+x2)24(x2+1)2(tan−1x)2(x2+1)

(x2+1)2y2+2x(x2+1)y′=4(x2+1)(tan⁡−1x)2(x2+1)2y2+2x(x2+1)y=4(x2+1)(tan−1x)2

(x2+1)2y2+2x(x2+1)y′=4(x2+1)y2(x2+1)2y2+2x(x2+1)y=4(x2+1)y2

Thus, we have verified that (x2+1)2y2+2x(x2+1)y′=2(x2+1)2y2+2x(x2+1)y=2.

 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability

Sadika

To verify Rolle's Theorem for the function y=x2+2y=x2+2 on the interval , we need to check three conditions: The function y=x2+2y=x2+2 is continuous on the closed interval . The function y=x2+2y=x2+2 is differentiable on the open interval (−2,2)(−2,2). The function takes the same value... read more

To verify Rolle's Theorem for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], we need to check three conditions:

  1. The function y=x2+2y=x2+2 is continuous on the closed interval [−2,2][−2,2].
  2. The function y=x2+2y=x2+2 is differentiable on the open interval (−2,2)(−2,2).
  3. The function takes the same value at the endpoints of the interval [−2,2][−2,2], i.e., y(−2)=y(2)y(−2)=y(2).

Let's check each condition:

  1. Continuity: The function y=x2+2y=x2+2 is a polynomial function and is continuous everywhere. Therefore, it is continuous on the closed interval [−2,2][−2,2].

  2. Differentiability: The function y=x2+2y=x2+2 is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the open interval (−2,2)(−2,2).

  3. Endpoints: y(−2)=(−2)2+2=4+2=6y(−2)=(−2)2+2=4+2=6 y(2)=(2)2+2=4+2=6y(2)=(2)2+2=4+2=6

    So, y(−2)=y(2)y(−2)=y(2).

Since all three conditions of Rolle's Theorem are satisfied, there exists at least one cc in the open interval (−2,2)(−2,2) such that f′(c)=0f(c)=0, where f(x)=x2+2f(x)=x2+2.

Now, let's find the derivative of y=x2+2y=x2+2:

y′=ddx(x2+2)=2xy=dxd(x2+2)=2x

To find the critical point(s), we set y′=0y=0: 2x=02x=0 x=0x=0

So, f′(0)=0f(0)=0.

Thus, Rolle's Theorem is verified for the function y=x2+2y=x2+2 on the interval [−2,2][−2,2], and there exists at least one point cc in the open interval (−2,2)(−2,2) where the derivative f′(c)f(c) is equal to zero.

 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability

Sadika

To differentiate sin⁡(2x)sin(2x) with respect to ecos⁡(x)ecos(x), we can use the chain rule. Let u=ecos⁡(x)u=ecos(x), then dudx=−esin⁡(x)dxdu=−esin(x). Now, we have: ddu(sin⁡(2x))=ddu(sin⁡(2x))×dudxdud(sin(2x))=dud(sin(2x))×dxdu Using the chain rule: ddu(sin⁡(2x))=dd(2x)(sin⁡(2x))×d(2x)dxdud(sin(2x))=d(2x)d(sin(2x))×dxd(2x) =cos⁡(2x)×2=cos(2x)×2 =2cos⁡(2x)=2cos(2x) Now,... read more

To differentiate sin⁡(2x)sin(2x) with respect to ecos⁡(x)ecos(x), we can use the chain rule. Let u=ecos⁡(x)u=ecos(x), then dudx=−esin⁡(x)dxdu=−esin(x).

Now, we have:

ddu(sin⁡(2x))=ddu(sin⁡(2x))×dudxdud(sin(2x))=dud(sin(2x))×dxdu

Using the chain rule:

ddu(sin⁡(2x))=dd(2x)(sin⁡(2x))×d(2x)dxdud(sin(2x))=d(2x)d(sin(2x))×dxd(2x)

=cos⁡(2x)×2=cos(2x)×2

=2cos⁡(2x)=2cos(2x)

Now, we need to multiply this by dudxdxdu:

=2cos⁡(2x)×(−esin⁡(x))=2cos(2x)×(−esin(x))

=−2ecos⁡(2x)sin⁡(x)=−2ecos(2x)sin(x)

So, the derivative of sin⁡(2x)sin(2x) with respect to ecos⁡(x)ecos(x) is −2ecos⁡(2x)sin⁡(x)−2ecos(2x)sin(x).

 
 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 06/04/2024 Learn Unit III: Calculus/Continuity and Differentiability

Sadika

To verify Rolle's Theorem for the function f(x)=x2+2x−8f(x)=x2+2x−8 on the interval , we need to check three conditions: f(x)f(x) is continuous on . f(x)f(x) is differentiable on (−4,2)(−4,2). f(−4)=f(2)f(−4)=f(2). Let's check these conditions one by one: Continuity... read more

To verify Rolle's Theorem for the function f(x)=x2+2x−8f(x)=x2+2x−8 on the interval [−4,2][−4,2], we need to check three conditions:

  1. f(x)f(x) is continuous on [−4,2][−4,2].
  2. f(x)f(x) is differentiable on (−4,2)(−4,2).
  3. f(−4)=f(2)f(−4)=f(2).

Let's check these conditions one by one:

  1. Continuity of f(x)f(x) on [−4,2][−4,2]:

The function f(x)=x2+2x−8f(x)=x2+2x−8 is a polynomial function and is continuous everywhere. Therefore, it is continuous on the interval [−4,2][−4,2].

  1. Differentiability of f(x)f(x) on (−4,2)(−4,2):

The function f(x)=x2+2x−8f(x)=x2+2x−8 is a polynomial function and is differentiable everywhere. Therefore, it is differentiable on the interval (−4,2)(−4,2).

  1. f(−4)=f(2)f(−4)=f(2):

Evaluate f(x)f(x) at the endpoints of the interval:

At x=−4x=−4: f(−4)=(−4)2+2(−4)−8=16−8−8=0f(−4)=(−4)2+2(−4)−8=16−8−8=0

At x=2x=2: f(2)=22+2(2)−8=4+4−8=0f(2)=22+2(2)−8=4+4−8=0

Since f(−4)=f(2)=0f(−4)=f(2)=0, Rolle's Theorem is satisfied on the interval [−4,2][−4,2].

Now, because f(x)f(x) is continuous on [−4,2][−4,2] and differentiable on (−4,2)(−4,2), and f(−4)=f(2)f(−4)=f(2), there must exist at least one point cc in the open interval (−4,2)(−4,2) such that f′(c)=0f(c)=0, which is the conclusion of Rolle's Theorem.

Therefore, Rolle's Theorem is verified for the function f(x)=x2+2x−8f(x)=x2+2x−8 on the interval [−4,2][−4,2].

 
 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 06/04/2024 Learn Unit III: Calculus/Integrals

Sadika

To evaluate the integral ∫3axb2+c2x2 dx∫b2+c2x23axdx we can use the substitution method. Let's make the substitution: u=cxu=cx Then, du=c dxdu=cdx, or dx=1c dudx=c1du. Substituting uu and dxdx into the integral: ∫3axb2+c2x2 dx=∫3ab2+u2⋅1c du∫b2+c2x23axdx=∫b2+u23a⋅c1du Now,... read more

To evaluate the integral

∫3axb2+c2x2 dxb2+c2x23axdx

we can use the substitution method. Let's make the substitution:

u=cxu=cx

Then, du=c dxdu=cdx, or dx=1c dudx=c1du.

Substituting uu and dxdx into the integral:

∫3axb2+c2x2 dx=∫3ab2+u2⋅1c dub2+c2x23axdx=b2+u23ac1du

Now, we can factor out the constant 3acc3a and rewrite the integral in a more familiar form:

3ac∫1b2+u2 duc3ab2+u21du

This is a standard integral, known to be the arctangent function:

3ac⋅1barctan⁡(ub)+Cc3ab1arctan(bu)+C

Now, we need to substitute back u=cxu=cx:

3ac⋅1barctan⁡(cxb)+Cc3ab1arctan(bcx)+C

Thus, the integral evaluates to:

3abarctan⁡(cxb)+Cb3aarctan(bcx)+C

where CC is the constant of integration.

 
 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Top topics in Class 12 Tuition

Looking for Class 12 Tuition ?

Find Online or Offline Class 12 Tuition on UrbanPro.

Do you offer Class 12 Tuition ?

Create Free Profile »

Looking for best Class 12 Tuition ?

POST YOUR REQUIREMENT
x

Ask a Question

Please enter your Question

Please select a Tag

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more