Ex 5.3, 11 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 11, 2021 by Teachoo
Last updated at March 11, 2021 by Teachoo
Transcript
Ex 5.3, 11 Find ππ¦/ππ₯ in, π¦ = cosβ1 ((1β π₯^2)/( 1+ π₯2 )) , 0 < x < 1 π¦ = cosβ1 ((1β π₯^2)/( 1+ π₯2 )) Putting x = tan ΞΈ y = γπππ γ^(β1) ((1βtanβ‘2 π)/(1+ tanβ‘2 π)) y = cosβ1 (cos 2π) π¦ =2ΞΈ Putting value of ΞΈ = γπ‘ππγ^(β1) π₯ π¦=2 (γπ‘ππγ^(β1) π₯) (cosβ‘2ΞΈ " =" (1 β tanβ‘2 π)/(1+ tanβ‘2 π)) Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ
Finding derivative of Inverse trigonometric functions
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 You are here
Ex 5.3, 10 Important
Ex 5.3, 15 Important
Misc 5 Important
Misc 4
Misc 13 Important
Finding derivative of Inverse trigonometric functions
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