Question

12.Find dy/dx of the functions given.
x^y + y^x = 1​

Answer 1

Answer:

x^y=x^y

ln x^y =ln x^y

x^y = e^(ln x^y) = e^(y lnx)

d(e^(y lnx))/dx

= e^(y lnx) d(y lnx)/dx

= e^(y lnx) (y d(lnx)/dx + lnx dy/dx)

= e^(y lnx) (y/x + lnx dy/dx)

= x^y (y/x + lnx dy/dx)

y^x=y^x

ln y^x = ln y^x

y^x = e^(ln y^x) = e^(x lny)

d(e^(x lny))/dx

= e^(x lny) d(x lny))/dx

= e^(x lny) (x d(lnx)/dx + lny)

= e^(x lny) (x(1/y)(dy/dx) + lny)

= y^x (x(1/y)(dy/dx) + lny)

x^y (y/x + lnx dy/dx) + y^x ((x/y)(dy/dx) + lny) =0