1.

If `(cosx)^y=(cosy)^x`find `(dy)/(dx)`.

Answer» `("cos" x)^(y) = ("cos" y)^(x)`
`rArr "log" ("cos" x)^(y) = "log" ("cos" y)^(x)`
`rArr y"log cos" x = x"log cos"y`
Differentiate both sides w.r.t.x
`y * (d)/(dx) "log cos"x + "log cos"x (d)/(dx)y`
`= x (d)/(dx)"log cos"y + "log cos"y(d)/(dx) x`
`rArr y * ((-"sin"x))/("cos"x) + "log cos"x * (dy)/(dx)`
`=(x(-"sin"y))/("cos"y)(dy)/(dx) + "log cos"y * 1`
`rArr -y"tan"x + "log cos"x (dy)/(dx)`
`= -x"tan"y (dy)/(dx) + "log cos"y`
`rArr ("log cos" x + x"tan"y)(dy)/(dx) = "log cos" y + y"tan"x`
`rArr (dy)/(dx) = ("log cos"y + y"tan"x)/("log cos"x + x"tan"y)`


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