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Misc 13 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at April 13, 2021 by Teachoo
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, 14
Ex 5.3, 9 Important
Ex 5.3, 13 Important
Ex 5.3, 12 Important
Ex 5.3, 11 Important
Ex 5.3, 10 Important
Ex 5.3, 15 Important
Misc 5 Important
Misc 4
Misc 13 Important You are here
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
Transcript
Misc 13 Find ππ¦/ππ₯ , if π¦=γπ ππγ^(βπ) π₯+γπ ππγ^(β1) β(1βπ₯2), β 1 β€ π₯ β€ 1 π¦=γπ ππγ^(βπ) π₯+γπ ππγ^(β1) β(1βπ₯^2 ) , β 1 β€ π₯ β€ 1 Putting π = πππβ‘π½ π¦=γπ ππγ^(βπ) (sinβ‘π)+γπ ππγ^(β1) β(1βsin^2 π ) π¦=π½+γπ ππγ^(β1) β(γππ¨π¬γ^π π ) π¦=π+γπ ππγ^(β1) (cos π) π¦=π+γπ ππγ^(β1) (sinβ‘(π /π βπ½) ) π¦=π+ (π/2 βπ) π¦=πβπ + π/2 π= π /π Differentiating π€.π.π‘.π₯. ππ¦/ππ₯ = π(π/2)/ππ₯ π π/π π = 0
