Question 20
If y = β(sinβ‘γπ₯+π¦γ ), then ππ¦/ππ₯ is equal to
(A) cosβ‘π₯/(2π¦β1) (B) cosβ‘π₯/(1β2π¦)
(C) sinβ‘π₯/(1β2π¦) (D) (β4π₯^3)/(2π¦ β1)
π¦=β(π ππβ‘γπ₯+π¦γ )
Squaring both sides
π¦^2=(β(sinβ‘γπ₯+π¦γ ))^2
π^π=πππβ‘γπ+πγ
π¦^2βπ¦=sinβ‘π₯
Differentiating both sides wrt π₯
(πγ(π¦γ^2))/ππ₯βππ¦/ππ₯ = (π(π ππβ‘γπ₯)γ)/ππ₯
ππ π π/π πβππ¦/ππ₯=πππβ‘π
ππ¦/ππ₯(2yβ1) =πππ β‘π₯
π π/π π = πππβ‘π/((ππ β π))
So, the correct answer is (A)

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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