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प्रश्न
Differentiate the following w.r.t. x : `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`
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उत्तर
Let y = `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`
= `"cosec"^-1(1/(cos6x))` ...[∵ cos3x = 4cos3x – 3cosx]
= cosec–1(sec6x)
= `"cosec"^-1["cosec"(pi/2 - 6x)]`
= `pi/(2) - 6x`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(pi/2 - 6x)`
= `"d"/"dx"(pi/2) - 6"d"/"dx"(x)`
= 0 – 6 x 1
= –6.
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