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प्रश्न
Differentiate the following w.r.t. x : `cos^-1((sqrt(3)cosx - sinx)/(2))`
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उत्तर
Let y = `cos^-1((sqrt(3)cosx - sinx)/(2))`
= `cos^-1[(cosx)((sqrt3)/2) - (sinx)(1/2)]`
= `cos^-1(cosx cos pi/6 - sinx sin pi/6) ...[∵ cos pi/6 = sqrt(3)/2, sin pi/6 = (1)/(2)]`
= `cos^-1[cos(x + pi/6)]`
= `x + pi/(6)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(x + pi/6)`
= `"d"/"dx"(x) + "d"/"dx"(pi/6)`
= 1 + 0
= 1.
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