Ex 5.4, 2 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 5.4, 2 (Method 1) Differentiate . . . x in , sin 1 Let = sin 1 Differentiating both sides . . . = sin 1 = sin 1 . sin 1 = sin 1 . 1 1 2 = Ex 5.4, 2 (Method 2) Differentiate . . . x in , sin 1 Let = sin 1 Let sin 1 = = Differentiating both sides . . . = We need in denominator, so multiplying & Dividing by . = = = = Putting value of = sin 1 sin 1 = sin 1 1 1 2 =
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.
