# Ex 5.3, 8 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.3, 8 Find in, sin2 + cos2 = 1 sin2 + cos2 = 1 Differentiating both sides . . . . sin2 + cos2 = 1 (sin2 ) + ( cos 2 ) = 0 Calculating Derivative of sin2 & cos 2 sepretaly Finding Derivative of (sin2 ) =2 2 1 . ( sin 2 ) =2 sin . ( sin ) =2 sin cos Finding Derivative of (cos2 ) =2 cos 2 1 . ( cos ) =2 cos . ( sin ) . ( ) = 2 cos sin . Now, (sin2 ) + (cos2 ) = 0 2 sin . cos + 2 cos sin . = 0 2 sin . cos 2 sin . cos . = 0 2 sin . cos . = 2 sin cos sin 2 . = sin 2 = sin 2 sin 2 =

Finding derivative of Implicit functions

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

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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.