Misc 6 - Differentiate cot-1 [root 1 + sin x + 1-sin x - Miscellaneous

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Misc 6 (Method 1) Differentiate w.r.t. x the function, 𝑐𝑜𝑡﷮−1 ﷯ ﷮1+ sin﷮𝑥﷯﷯ + ﷮1− sin﷮𝑥﷯﷯﷮ ﷮1+ sin﷮𝑥 ﷯﷯− ﷮1− sin﷮𝑥﷯﷯﷯ ﷯ , 0<𝑥< 𝜋﷮2﷯ Let 𝑦= 𝑐𝑜𝑡﷮−1 ﷯ ﷮1+ sin﷮𝑥﷯﷯ + ﷮1− sin﷮𝑥﷯﷯﷮ ﷮1+ sin﷮𝑥 ﷯﷯− ﷮1− sin﷮𝑥﷯﷯﷯ ﷯ Rationalizing the sum 𝑦= 𝑐𝑜𝑡﷮−1 ﷯ ﷮1+ sin﷮𝑥﷯﷯ + ﷮1− sin﷮𝑥﷯﷯﷯﷮ ﷮1+ sin﷮𝑥 ﷯﷯− ﷮1− sin﷮𝑥﷯﷯﷯﷯ × ﷮1+ sin﷮𝑥﷯﷯ + ﷮1− sin﷮𝑥﷯﷯﷯﷮ ﷮1+ sin﷮𝑥 ﷯﷯+ ﷮1− sin﷮𝑥﷯﷯﷯﷯﷯ 𝑦= 𝑐𝑜𝑡﷮−1 ﷯ ﷮1 + sin﷮𝑥﷯﷯ + ﷮1 − sin﷮𝑥﷯﷯﷯﷮2﷯﷮ ﷮1 + sin﷮𝑥 ﷯﷯− ﷮1 − sin﷮𝑥﷯﷯﷯ ﷮1+ sin﷮𝑥 ﷯﷯+ ﷮1− sin﷮𝑥﷯﷯﷯﷯ ﷯ = 𝑐𝑜𝑡﷮−1 ﷯ ﷮1 + sin﷮𝑥﷯﷯ ﷯﷮2﷯+ ﷮1 − sin﷮𝑥﷯﷯ ﷯﷮2﷯+2 ﷮1 + sin﷮𝑥 ﷯﷯﷯ ﷮1 − sin﷮𝑥﷯﷯﷯﷮ ﷮1 + sin﷮𝑥 ﷯﷯− ﷮1 − sin﷮𝑥﷯﷯﷯ ﷮1 + sin﷮𝑥 ﷯﷯+ ﷮1 − sin﷮𝑥﷯﷯﷯﷯ ﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 1 + sin﷮𝑥﷯﷯ + 1 − sin﷮𝑥﷯﷯ + 2 ﷮ 1 + sin﷮𝑥﷯﷯ 1 − sin﷮𝑥﷯﷯﷯﷮ ﷮1 + sin﷮𝑥 ﷯﷯﷯﷮2﷯ − ﷮1 − sin﷮𝑥﷯﷯﷯﷮2﷯﷯ ﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 1 + sin﷮𝑥﷯ + 1 − sin﷮𝑥﷯ + 2 ﷮ 1﷯﷮2﷯ − sin﷮2﷯﷮𝑥﷯﷯﷮1 + sin﷮𝑥﷯ − 1 + sin﷮𝑥﷯﷯ ﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 2 + 2 ﷮1 − sin﷮2﷯﷮𝑥﷯﷯﷮2 sin﷮𝑥﷯﷯ ﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 2 1 + ﷮1 − sin﷮2﷯﷮𝑥﷯﷯ ﷯ ﷮2 sin﷮𝑥﷯﷯﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 1 + ﷮ c𝑜𝑠﷮2﷯﷮𝑥﷯﷯ ﷮ sin﷮𝑥﷯﷯﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 1 + cos﷮𝑥﷯ ﷮ sin﷮𝑥﷯﷯﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 1 + 2 cos﷮2﷯﷮ 𝑥﷮2﷯﷯ − 1 ﷮2 sin﷮ 𝑥﷮2﷯ ﷯ cos ﷮ 𝑥﷮2﷯﷯﷯﷯ = 𝑐𝑜𝑡﷮−1 ﷯ 2 cos﷮2﷯﷮ 𝑥﷮2﷯﷯ ﷮2 sin﷮ 𝑥﷮2﷯ ﷯ cos ﷮ 𝑥﷮2﷯﷯﷯﷯ = 𝑐𝑜𝑡﷮−1 ﷯ cos ﷮ 𝑥﷮2﷯﷯ ﷮ sin﷮ 𝑥﷮2﷯ ﷯﷯﷯ = 𝑐𝑜𝑡﷮−1 ﷯ cot﷮ 𝑥﷮2﷯﷯﷯﷯ = 𝑥﷮2﷯ Thus, 𝒚= 𝒙﷮𝟐﷯ Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮2﷯ 𝑑𝑥﷮𝑑𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝟏﷮𝟐﷯ Misc 6 (Method 2 ) Differentiate w.r.t. x the function, 𝑐𝑜𝑡﷮−1 ﷯ ﷮1+ sin﷮𝑥﷯﷯ + ﷮1− sin﷮𝑥﷯﷯﷮ ﷮1+ sin﷮𝑥 ﷯﷯− ﷮1− sin﷮𝑥﷯﷯﷯ ﷯ , 0<𝑥< 𝜋﷮2﷯ Let 𝑦= 𝑐𝑜𝑡﷮−1 ﷯ ﷮1+ sin﷮𝑥﷯﷯ + ﷮1− sin﷮𝑥﷯﷯﷮ ﷮1+ sin﷮𝑥 ﷯﷯− ﷮1− sin﷮𝑥﷯﷯﷯ ﷯ As 𝑠𝑖𝑛﷮2﷯𝜃+ 𝑐𝑜𝑠﷮2﷯﷮𝜃﷯=1 Replacing 𝜃 by 𝑥﷮2﷯ 𝑠𝑖𝑛2 𝑥﷮2﷯ + 𝑐𝑜𝑠﷮2﷯ 𝑥﷮2﷯ = 1 & 𝑠𝑖𝑛﷮2𝜃﷯=2 𝑠𝑖𝑛﷮𝜃 𝑐𝑜𝑠﷮𝜃﷯﷯ Replacing 𝜃 by 𝑥﷮2﷯ 𝑠𝑖𝑛⁡𝑥 = 2 𝑠𝑖𝑛⁡ 𝑥﷮2﷯ 𝑐𝑜𝑠⁡ 𝑥﷮2﷯ Now, y = 𝑐𝑜𝑠﷮−1﷯ ﷮1 + sin﷮𝑥﷯﷯ + ﷮1 − sin﷮𝑥﷯﷯﷮ ﷮1 + sin﷮𝑥﷯﷯ − ﷮1 − sin﷮𝑥﷯﷯﷯﷯ Substitute value of ﷮1+ 𝑠𝑖𝑛﷮𝑥﷯﷯ & ﷮1− 𝑠𝑖𝑛﷮𝑥﷯﷯ from (1) & (2). y = cot−1 cos ﷮ 𝑥﷮2﷯﷯ + sin ﷮ 𝑥﷮2﷯﷯﷯ + cos ﷮ 𝑥﷮2﷯﷯ − sin ﷮ 𝑥﷮2﷯﷯﷯﷮ cos ﷮ 𝑥﷮2﷯﷯ + sin ﷮ 𝑥﷮2﷯﷯﷯ + cos ﷮ 𝑥﷮2﷯﷯ − sin ﷮ 𝑥﷮2﷯﷯﷯﷯﷯ y = cot−1 cos ﷮ 𝑥﷮2﷯﷯ + sin ﷮ 𝑥﷮2﷯﷯﷯ + cos ﷮ 𝑥﷮2﷯﷯ − sin ﷮ 𝑥﷮2﷯﷯﷯﷮ cos ﷮ 𝑥﷮2﷯﷯ + sin ﷮ 𝑥﷮2﷯﷯﷯ + cos ﷮ 𝑥﷮2﷯﷯ − sin ﷮ 𝑥﷮2﷯﷯﷯﷯﷯ 𝑦= 𝑐𝑜𝑡﷮−1 ﷯ 𝑐𝑜𝑠﷮ 𝑥﷮2﷯﷯ + 𝑠𝑖𝑛 ﷮ 𝑥﷮2﷯﷯ + 𝑐𝑜𝑠﷮ 𝑥﷮2﷯﷯ − 𝑠𝑖𝑛 ﷮ 𝑥﷮2﷯﷯﷮ 𝑐𝑜𝑠﷮ 𝑥﷮2﷯﷯ + 𝑠𝑖𝑛﷮ 𝑥﷮2﷯﷯ − 𝑐𝑜𝑠 ﷮ 𝑥﷮2﷯﷯ − 𝑠𝑖𝑛﷮ 𝑥﷮2﷯﷯﷯ ﷯ 𝑦= 𝑐𝑜𝑡﷮−1 ﷯ 2 𝑐𝑜𝑠 ﷮ 𝑥﷮2﷯﷯ ﷮2 𝑠𝑖𝑛﷮ 𝑥﷮2﷯﷯﷯ ﷯ 𝑦= 𝑐𝑜𝑡﷮−1 ﷯ 𝑐𝑜𝑡 𝑥 ﷮2﷯ ﷯ 𝑦= 𝑥﷮2﷯ Thus, 𝒚= 𝒙﷮𝟐﷯ Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑﷮𝑑𝑥﷯ 𝑥﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 1﷮2﷯ 𝑑𝑥﷮𝑑𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝟏﷮𝟐﷯

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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.