If  y = √(sin⁡γ€–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



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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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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.