Misc 14 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at May 29, 2018 by Teachoo
Transcript
Misc 14 If 1+ + 1+ = 0 , for 1 < < 1, prove that = 1 1 + 2 1+ + 1+ = 0 1+ = 1+ Squaring both sides 1+ 2 = 1+ 2 2 1+ 2 = 2 1+ 2 2 1+ = 2 1+ 2 + 2 = 2 + 2 2 2 = xy2 x2y + = + = + = + = = = + = +1 + 1 = = + 1 Differentiating . . . . = + 1 = + 1 + 1 8 . + 1 2 = 1 + 1 1 + 0 + 1 2 = 1 + 1 2 = 1 + 1 2 Hence, = +
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.