Ex 5.3, 11 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Sept. 17, 2019 by Teachoo
Transcript
Ex 5.3, 11 Find 𝑑𝑦𝑑𝑥 in, 𝑦 = cos–1 1− 𝑥2 1+ 𝑥2 , 0 < x < 1 𝑦 = cos–1 1− 𝑥2 1+ 𝑥2 Putting x = tan θ y = 𝑐𝑜𝑠−1 1− tan2𝜃1+ tan2 𝜃 y = cos−1 (cos 2𝜃) 𝑦 =2θ Putting value of θ = 𝑡𝑎𝑛−1 𝑥 𝑦=2 ( 𝑡𝑎𝑛−1 𝑥) Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . 𝑑(𝑦)𝑑𝑥 = 𝑑 (2 𝑡𝑎𝑛−1 𝑥 ) 𝑑𝑥 𝑑𝑦𝑑𝑥 = 2 𝑑 ( 𝑡𝑎𝑛−1 𝑥 ) 𝑑𝑥 𝑑𝑦𝑑𝑥 = 2 11+ 𝑥2 𝒅𝒚𝒅𝒙 = 𝟐𝟏+ 𝒙𝟐
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)
Derivative of sec-1 x (Sec inverse x)
Derivative of cosec-1 x (Cosec inverse x)
Ex 5.3, 9
Ex 5.3, 12 Important
Ex 5.3, 14 Important
Ex 5.3, 11 Important You are here
Ex 5.3, 13
Ex 5.3, 10 Important
Ex 5.3, 15 Important
Misc 5 Important
Misc 4
Misc 13
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.