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Misc 1 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at April 13, 2021 by Teachoo
Transcript
Misc 1 Differentiate w.r.t. x the function, (3๐ฅ2 โ 9๐ฅ + 5)^9Let ๐ฆ=(3๐ฅ2 โ 9๐ฅ + 5)^9 Differentiating ๐ค.๐.๐ก.๐ฅ. ๐๐ฆ/๐๐ฅ = (๐(3๐ฅ2 โ 9๐ฅ + 5)^9)/๐๐ฅ = 9(3๐ฅ2 โ 9๐ฅ + 5)^(9 โ1) . ๐(3๐ฅ2 โ 9๐ฅ + 5)/๐๐ฅ = 9(3๐ฅ2 โ 9๐ฅ + 5)^8 . (๐(3๐ฅ2)/๐๐ฅ โ ๐(9๐ฅ)/๐๐ฅ โ ๐(5)/๐๐ฅ) = 9(3๐ฅ2 โ 9๐ฅ + 5)^8 . (6๐ฅโ9+0) = 9(3๐ฅ2 โ 9๐ฅ + 5)^8 . (6๐ฅโ9) = 9(3๐ฅ2 โ 9๐ฅ + 5)^8 . 3(2๐ฅโ3) = ๐๐(๐๐๐ โ ๐๐ + ๐)^๐ (๐๐โ๐)
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Finding derivative of a function by chain rule
Chapter 5 Class 12 Continuity and Differentiability (Term 1)
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
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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 10 years. He provides courses for Maths and Science at Teachoo.