If `"x" = "e"^(cos2"t") "and" "y" = "e"^(sin2"t")`, prove that `(d"y")/(d"x") = - ("y"log"x")/("x"log"y")`.

#### Solution

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")`