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


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Example 40 (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 40 (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

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.