Ex 5.2, 2 - Differentiate cos (sin⁑ x) - Chapter 5 NCERT - Finding derivative of a function by chain rule

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.2, 2 Differentiate the functions with respect to π‘₯ cos (sin⁑π‘₯) Let 𝑦 = cos (sin⁑π‘₯) We need to find derivative of 𝑦, 𝑀.π‘Ÿ.𝑑.π‘₯ i.e. 𝑦′ = (cos (sin⁑π‘₯ ))’ (𝑑𝑦 )/𝑑π‘₯ = (𝑑(cos (sin⁑π‘₯ )))/𝑑π‘₯ = βˆ’ sin⁑(sin⁑π‘₯) . (𝑑(sin⁑〖π‘₯)γ€—)/𝑑π‘₯ = βˆ’ sin⁑(sin⁑π‘₯) . cos⁑π‘₯ = βˆ’ πœπ¨π¬β‘π’™ π’”π’Šπ’ ⁑(𝐬𝐒𝐧⁑〖𝒙)γ€—

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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.