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Example 45 - Differentiate (i) cos-1 (sin x) - Chapter 5

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Example 45 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 45 (Method 1) Differentiate the following 𝑀.π‘Ÿ.𝑑. π‘₯. (i) cos^(βˆ’1) (sin⁑π‘₯) Let 𝑓(π‘₯) = cos^(βˆ’1) (sin⁑π‘₯) 𝑓(π‘₯) = cos^(βˆ’1) (γ€–cos 〗⁑(πœ‹/2 βˆ’π‘₯) ) 𝒇(𝒙) = 𝝅/𝟐 βˆ’π’™ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑓’(π‘₯) = (𝑑 (πœ‹/2))/𝑑π‘₯ βˆ’ (𝑑(π‘₯))/𝑑π‘₯ 𝑓’(π‘₯) = 0 βˆ’ 1 𝒇’(𝒙) = βˆ’ 1 (𝐴𝑠 γ€– 𝑠𝑖𝑛 πœƒ 〗⁑〖=γ€–π‘π‘œπ‘  〗⁑〖(πœ‹/2 βˆ’π‘₯)γ€— γ€— ) ("As " (𝑑(π‘₯))/𝑑π‘₯ " = 1 & " πœ‹/2 " is constant" ) Example 45 (Method 2) Differentiate the following 𝑀.π‘Ÿ.𝑑. π‘₯. (i) cos^(βˆ’1) (sin⁑π‘₯) Let 𝑓(π‘₯) = cos^(βˆ’1) (sin⁑π‘₯) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑓′(π‘₯) = (βˆ’1)/√(1 βˆ’ γ€–(sin⁑π‘₯)γ€—^2 ) Γ— (sin⁑π‘₯ )^β€² 𝑓′(π‘₯) = (βˆ’1)/√(1 βˆ’ sin^2⁑π‘₯ ) Γ—cos⁑π‘₯ 𝑓′(π‘₯) = (βˆ’1)/√(cos^2⁑π‘₯ ) Γ—cos⁑π‘₯ 𝑓′(π‘₯) = (βˆ’1)/cos⁑π‘₯ Γ—cos⁑π‘₯ 𝒇’(𝒙) = βˆ’1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.