Question

Given that x^y + y^x = a^b, where a and b are positive constants, find `dy/dx`.

Answer 1

Given, x^y + y^x = a^b

`e^(logx^y) + e^(logy^x) = a^b`   ...[we know that: `e^(log m) = m]`

`e^(ylogx) + e^(x log y) = a^b`   ...[log mⁿ = n log m]

d.w.r to x

`e^(y log x) [yxx1/x + log x dy/dx]+ e^(xlogy)[x xx 1/y dy/dx + log y xx 1] = d/dx a^b`

`e log x^y [y/x + logx dy/dx] + e^(logy^x) [x/y dy/dx + logy] = 0`

`x^y[y/x + log x dy/dx] + y^x[x/y dy/dx + logy] = 0`

`yx^(y-1) + x^y log x dy/dx + xy^(x-1) dy/dx + y^xlogy = 0`

`(x^y logx+xy^(x-1))dy/dx = -(y^x logy+yx^(y-1))`

`dy/dx = (-y^x logy+yx^(y-1))/(x^y logx+xy^(x-1)`