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Question
Find `"dy"/"dx"` if : x = cosec2θ, y = cot3θ at θ= `pi/(6)`
Solution
x = cosec2θ, y = cot3θ
Differentiating x and y w.r.t. θ, we get
`"dx"/"dθ" = "d"/"dθ"("cosec"θ)^2 = 2"cosec"θ."d"/"dθ"("cosec"θ)`
= 2cosecθ(– cosecθ cotθ)
= – 2cosec2θ cotθ
and
`"dy"/"dθ" = "d"/"dθ"(cotθ)^3 = 3cot^2θ."d"/"dθ"(cotθ)`
= 3cot2θ.(–cosec2θ)
= –3cot2θ.cosec2θ
∴ `"dy"/"dx" = (("dy"/"dθ"))/(("dx"/"dθ")) = (-3cot^2θ."cosc"^2θ)/(-2"cosec"^2θ.cotθ)`
= `(3)/(2)cotθ`
∴ `(dy/dx)_("at" θ = pi/6)`
= `(3)/(2)cot pi/(6)`
= `(3sqrt(3))/(2)`.
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