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प्रश्न
Differentiate the following w.r.t.x: `cot(logx/2) - log(cotx/2)`
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उत्तर
Let y = `cot(logx/2) - log(cotx/2)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[cot(logx/2) - log(cotx/2)]`
= `"d"/"dx"[cot(logx/2)] - "d"/"dx"[log(cotx/2)]`
= `-"cosec"^2(logx/2)."d"/"dx"(logx/2) - (1)/((cotx/2))."d"/"dx"(cotx/2)`
= `-"cosec"^2(logx/2) xx 1/2 xx 1/x - 2/cotx xx 1/2 xx (-"cosec"^2x)`
= `-("cosec"^2(logx/2))/(2x) + tanx."cosec"^2x`.
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