Sum

**Solve the following problem.**

Obtain derivative of the following function: `"x"/"sin x"`

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#### Solution

Using `"d"/"dx" [("f"_1("x"))/("f"_2("x"))] = 1/("f"_2("x")) ("df"_1("x"))/"dx" - ("f"_1("x"))/("f"_2^2("x")) ("df"_2("x"))/"dx"`

For f_{1}(x) = x and f_{2}(x) = sin x

∴ `"d"/"dx"("x"/"sin x") = 1/"sin x" xx ("d"("x"))/"dx" - "x"/sin^2"x" xx ("d"(sin "x"))/"dx"`

`= 1/"sin x" xx 1 - "x"/("sin"^2"x") xx cos "x" .....[because "d"/"dx" ("sin x") = "cos x"]`

`= 1/"sin x" - "x cos x"/sin^2"x"`

Concept: Introduction to Calculus

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