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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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