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Question
Differentiate `x/sinx` w.r.t. sin x
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Solution
Let y = `x/sinx` and z = sin x.
Differentiating both the parametric functions w.r.t. x
`"dy"/"dx" = (sin x * "d"/"dx" (x) - x * "d"/"dx" (sin x))/(sin x)^2`
= `(sin x * 1 - x * cos x)/(sin^2x)`
= `(sinx - x cos x)/(sin^2x)`
`"dz"/"dx"` = cos x
∴ `"dy"/"dz" = ("dy"/"dx")/("dz"/"dx")`
= `((sinx - x cos x)/sin^2x)/cosx`
= `(sinx - xcosx)/(sin^2x cos x)`
= `sinx/(sin^2x cosx) - (xcosx)/(sin^2x cosx)`
= `tanx/(sin^2x) - x/(sin^2x)`
= `(tanx - x)/(sin^2x)`
Hence, `"dy"/"dz" = (tanx - x)/(sin^2x)`.
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