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