Misc 4 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at December 16, 2024 by Teachoo
Finding derivative of Inverse trigonometric functions
Derivative of cos-1 x (Cos inverse x)
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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
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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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Transcript
Misc 4 Differentiate š¤.š.š”. š„ the function, ćš ššć^(ā1) (š„ āš„), 0 ā¤ š„ ā¤ 1 Let š¦=ćš ššć^(ā1) (š„ āš„) š¦=ćš ššć^(ā1) (š„ . š„^(1/2)) š¦=ćš ššć^(ā1) (š„^(1 + 1/2)) š=ćšššć^(āš) (š^( š/š) ) Differentiating š¤.š.š”. š„ šš¦/šš„ = š(ćš ššć^(ā1) (š„^( 3/2)))/šš„ šš¦/šš„ = 1/ā(1 ā (š„^(3/2) )^2 ) Ć (š(š„)^(3/2))/šš„ ("As " š(ćš ššć^(ā1)ā”š„ )/šš„=1/ā(1 ā š„^2 )) šš¦/šš„ = 1/ā(1 ā š„^3 ) Ć 3/2 (š„)^(3/2 ā1) = 1/ā(1 ā š„^3 ) Ć 3/2 ćš„ ć^(1/2 ) = 1/ā(1 ā š„^3 ) Ć 3/2 āš„ = š/š ā(š/(š āš^š ))
