Question 20 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability

Last updated at August 12, 2026 by

IfΒ  y = √(sin⁑〖x+yγ€— ), then dy/dx is equal to

(A) cos⁑x/(2y-1) 

(B) cos⁑x/(1-2y)

(C) sin⁑x/(1-2y) 

(D) (-4x 3 )/(2y -1)

This question is similar to Example 25 - Chapter 5 Class 12 - Continuity and Differentiability

MCQ Class 12 - If y = √sin⁑ (π‘₯ + 𝑦), then 𝑑𝑦/𝑑π‘₯ is equal to - NCERT Exemplar - MCQs

part 2 - Question 20 - NCERT Exemplar - MCQs - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability

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Transcript

Question 20 If y = √(sin⁑〖π‘₯+𝑦〗 ), then 𝑑𝑦/𝑑π‘₯ is equal to (A) cos⁑π‘₯/(2π‘¦βˆ’1) (B) cos⁑π‘₯/(1βˆ’2𝑦) (C) sin⁑π‘₯/(1βˆ’2𝑦) (D) (βˆ’4π‘₯^3)/(2𝑦 βˆ’1) 𝑦=√(𝑠𝑖𝑛⁑〖π‘₯+𝑦〗 ) Squaring both sides 𝑦^2=(√(sin⁑〖π‘₯+𝑦〗 ))^2 π’š^𝟐=π’”π’Šπ’β‘γ€–π’™+π’šγ€— 𝑦^2βˆ’π‘¦=sin⁑π‘₯ Differentiating both sides wrt π‘₯ (𝑑〖(𝑦〗^2))/𝑑π‘₯βˆ’π‘‘π‘¦/𝑑π‘₯ = (𝑑(𝑠𝑖𝑛⁑〖π‘₯)γ€—)/𝑑π‘₯ πŸπ’š π’…π’š/π’…π’™βˆ’π‘‘π‘¦/𝑑π‘₯=𝒄𝒐𝒔⁑𝒙 𝑑𝑦/𝑑π‘₯(2yβˆ’1) =π‘π‘œπ‘ β‘π‘₯ π’…π’š/𝒅𝒙 = 𝒄𝒐𝒔⁑𝒙/((πŸπ’š βˆ’ 𝟏)) So, the correct answer is (A)

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