If x^ y + y ^x= a ^ b , then dy d x at x=1, y =2 is

Step-by-step solution

Differentiate the equation x^y + y^x = a^b with respect to x . Since a^b is a constant, the derivative is zero, yielding: d dx (x^y) + d dx (y^x) = 0 . Let u = x^y . Applying logarithmic differentiation: u = y x . Differentiate both sides: 1 u du dx = dy dx x + y x So, du dx = x^y ( dy dx x + y x ) . Similarly, for v = y^x : v = x y . Differentiate: 1 v dv dx = y + x y dy dx Thus, dv dx = y^x ( y + x y dy dx ) . Substitute into the derivative equation: x^y ( dy dx x + y x ) + y^x ( y + x y dy dx ) = 0 . Evaluate at