Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Last updated at Jan. 16, 2020 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - https://you.tube/Chapter-5-Class-12-Continuity

Transcript

Find the derivative of f given by f (x) = cot–1 𝑥 assuming it exists. Let y = cot–1 𝑥 co𝑡𝑦= 𝑥 𝑥 =co𝑡𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 (𝑑(𝑥))/𝑑𝑥 = (𝑑 (co𝑡𝑦 ))/𝑑 1 = (𝑑 (co𝑡𝑦 ))/𝑑𝑥 We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = (𝑑 (co𝑡𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = −cosec^2𝑦 . 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 (−cosec^2𝑦 ) = 1 𝑑𝑦/𝑑𝑥 = (−1)/(〖𝒄𝒐𝒔𝒆𝒄〗^𝟐 𝒚) 𝑑𝑦/𝑑𝑥 = (−1)/(𝟏 + 〖𝐜𝐨𝐭〗^𝟐𝒚) (𝐴𝑠 〖 𝑐𝑜𝑠𝑒𝑐〗^2〖𝑦= 〖1+〗〖𝑐𝑜𝑡〗^2𝑦 〗) Putting 𝑐𝑜𝑡𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = (−1)/(𝒙^𝟐 + 1) As 𝑦 = cot^(−1) 𝑥 So, 𝑐𝑜𝑡𝑦 = 𝑥 Hence, (𝒅(〖𝐜𝐨𝒕〗^(−𝟏)〖𝒙)〗)/𝒅𝒙 = (−𝟏)/(𝒙^𝟐 + 𝟏)

Finding derivative of Inverse trigonometric functions

Derivative of cos-1 x (Cos inverse x)

Example 26 Important

Example 27

Derivative of cot-1 x (cot inverse x) You are here

Derivative of sec-1 x (Sec inverse x)

Derivative of cosec-1 x (Cosec inverse x)

Ex 5.3, 14

Ex 5.3, 9 Important

Ex 5.3, 13 Important

Ex 5.3, 12 Important

Ex 5.3, 11 Important

Ex 5.3, 10 Important

Ex 5.3, 15 Important

Misc 5 Important

Misc 4

Misc 13 Important

Chapter 5 Class 12 Continuity and Differentiability

Concept wise

- Checking continuity at a given point
- Checking continuity at any point
- Checking continuity using LHL and RHL
- Algebra of continous functions
- Continuity of composite functions
- Checking if funciton is differentiable
- Finding derivative of a function by chain rule
- Finding derivative of Implicit functions
- Finding derivative of Inverse trigonometric functions
- Finding derivative of Exponential & logarithm functions
- Logarithmic Differentiation - Type 1
- Logarithmic Differentiation - Type 2
- Derivatives in parametric form
- Finding second order derivatives - Normal form
- Finding second order derivatives- Implicit form
- Proofs
- Verify Rolles theorem
- Verify Mean Value Theorem

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.