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Misc 4 - 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 You are here
Misc 13 Important
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
Transcript
Misc 4 Differentiate π€.π.π‘. π₯ the function, γπ ππγ^(β1) (π₯ βπ₯), 0 β€ π₯ β€ 1 Let π¦=γπ ππγ^(β1) (π₯ βπ₯) π¦=γπ ππγ^(β1) (π₯ . π₯^(1/2)) π¦=γπ ππγ^(β1) (π₯^(1 + 1/2)) π=γπππγ^(βπ) (π^( π/π) ) Differentiating π€.π.π‘. π₯ ππ¦/ππ₯ = π(γπ ππγ^(β1) (π₯^( 3/2)))/ππ₯ ππ¦/ππ₯ = 1/β(1 β (π₯^(3/2) )^2 ) Γ (π(π₯)^(3/2))/ππ₯ ("As " π(γπ ππγ^(β1)β‘π₯ )/ππ₯=1/β(1 β π₯^2 )) ππ¦/ππ₯ = 1/β(1 β π₯^3 ) Γ 3/2 (π₯)^(3/2 β1) = 1/β(1 β π₯^3 ) Γ 3/2 γπ₯ γ^(1/2 ) = 1/β(1 β π₯^3 ) Γ 3/2 βπ₯ = π/π β(π/(π βπ^π ))
