#### Question

Sum

If `(sin "x")^"y" = "x" + "y", "find" (d"y")/(d"x")`

#### Solution

`(sin "x")^"y" = "x" + "y"`

Take log on both the sides,

`log(sin "x")^"y" = log("x" + "y")`

⇒ `"y" log (sin "x") = log ("x" + "y")` ......(i)

Differentiate (i) w.r.t.x

`log (sin "x")· (d"y")/(d"x") + "y"· (d)/(d"x") [ log(sin "x")] = (d)/(d"x") [log ("x"+"y") ]`

⇒ `log (sin "x")· (d"y")/(d"x") + "y"· (cos "x")/(sin"x") = (1)/(("x"+"y"))· (1+ (d"y")/(d"x"))`

⇒ `(d"y")/(d"x") [ log( sin "x") - (1)/(("x"+"y"))] = (1)/(("x"+"y")) - "y"·cot "x" `

⇒ `(d"y")/(d"x") = (1 - ("xy" + "y"^2)·cot "x")/(("x"+"y")·log (sin "x") -1)`

Concept: Logarithmic Differentiation

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