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प्रश्न
If `"x" = "e"^(cos2"t") "and" "y" = "e"^(sin2"t")`, prove that `(d"y")/(d"x") = - ("y"log"x")/("x"log"y")`.
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उत्तर
Here `"x" = "e"^(cos2"t") and "y"="e"^(sin2"t")`
⇒ `log_"e" "x" = log_"e" ("e"^(cos2"t")) and log_"e" "y" = log_"e"("e"^(sin2"t"))`
⇒ `log_"e" "x" = cos 2"t" log_"e" ("e") and log_"e" "y"=sin 2"t" log_"e" ("e") ...["As log"_"e" ("e") = 1]`
∴ `log_"e" "x" = cos 2"t" and log_"e" "y" = sin 2"t"`
Squaring and then adding these two equations,
`(log_"e" "x")^2 + (log_"e" "y")^2 = cos^2 2"t" + sin^2 2"t"`
⇒ `(log_"e" "x")^2 + (log_"e" "y")^2 = 1`
⇒ `2(log_"e" "x") xx (1)/("x") + 2(log_"e" "y") xx (1)/("y") xx (d"y")/(d"x") = 0`
∴ `(d)/(d"x") [(log_"e" "x")^2 + (log_"e" "y")^2] = (d)/(d"x") ...(1)`
⇒ `(log_"e" "y")/("y") xx (d"y")/(d"x") = - (log_"e" "x")/("x")`
∴ `(d"y")/(d"x") = - ("y"log"x")/("x"log"y")`
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