If š‘¦ = sin -1 x, then (1-x 2 ) y 2 š‘–š‘  equal to

(a) xy 1 Ā 

(b) š‘„š‘¦Ā 

(c) xy 2 Ā 

(d) x 2

Ā 

[Sample Paper MCQ If y = sin^-1 x, then (1 - x^2)y2 is equal to - CBSE Class 12 Sample Paper for 2023 Boards

part 2 - Question 16 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 16 If š‘¦ = ć€–š‘ š‘–š‘›ć€—^(āˆ’1) š‘„, then (1āˆ’š‘„^2 ) š‘¦_2 š‘–š‘  equal to (a) ć€–š‘„š‘¦ć€—_1 (b) š‘„š‘¦ (c) ć€–š‘„š‘¦ć€—_2 (d) š‘„^2 Given š‘¦ = ć€–š‘ š‘–š‘›ć€—^(āˆ’1) š‘„ Differentiating w.r.t x š‘‘š‘¦/š‘‘š‘„=š‘‘(sin^(āˆ’1)ā”š‘„ )/š‘„ š’…š’š/š’…š’™=šŸ/√(šŸ āˆ’ š’™^šŸ ) š‘¦^′=1/√(1 āˆ’ š‘„^2 ) š’š^′ Ɨ √(šŸ āˆ’ š’™^šŸ )=šŸ Again differentiating w.r.t x (š‘¦^′ Ɨ √(1 āˆ’ š‘„^2 ))^′=š‘‘(1)/š‘‘š‘„ (š‘¦^′ Ɨ √(1 āˆ’ š‘„^2 ))^′=0 ć€–š’š^′′ √(šŸ āˆ’ š’™^šŸ )+š’š^′ (√(šŸ āˆ’ š’™^šŸ ))怗^′=šŸŽ ć€–š‘¦^′′ √(1 āˆ’ š‘„^2 )+š‘¦^′ Ɨ1/(2√(1 āˆ’ š‘„^2 )) (1 āˆ’ š‘„^2 )怗^′=0 š‘¦^′′ √(1 āˆ’ š‘„^2 )+š‘¦^′ Ɨ1/(2√(1 āˆ’ š‘„^2 )) Ɨ āˆ’2š‘„=0 š‘¦^′′ √(1 āˆ’ š‘„^2 )āˆ’š‘¦^′ Ć—š‘„/√(1 āˆ’ š‘„^2 ) =0 š‘¦^′′ √(1 āˆ’ š‘„^2 )=š‘¦^′ Ć—š‘„/√(1 āˆ’ š‘„^2 ) š‘¦^′′ √(1 āˆ’ š‘„^2 ) Ć—āˆš(1 āˆ’ š‘„^2 )=š‘¦^′ š‘„ š’š^′′ (šŸ āˆ’ š’™^šŸ) =š’š^′ š’™ So, the correct answer is (a)

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