Question

If x^y×y^x=M where M is a constant then find its
first derivatve

Answer 1

`x^y y^x=M`

`\ln(x^y y^x)=\ln M`

`y\ln x+x\ln y=\ln M`

Differentiating both sides with respect to
`x`, and assuming
`y=y(x)`, we have


`(\mathrm d)/(\mathrm dx)[y\ln x+x\ln y]=0`

`(\mathrm dy)/(\mathrm dx)\ln x+\frac yx+\ln y+\frac xy(\mathrm dy)/(\mathrm dx)=0`

`(\mathrm dy)/(\mathrm dx)\left(\ln x+\frac xy\right)=-\frac yx-\ln y`

`(\mathrm dy)/(\mathrm dx)=-(\frac yx+\ln y)/(\frac xy+\ln x)`

`(\mathrm dy)/(\mathrm dx)=-(y^2+xy\ln y)/(x^2+xy\ln x)`

In case you wanted the derivative with respect to
`y` under the assumption that
`x=x(y)`, the result would be the same with the exception that the left hand side getting written as
`(\mathrm dx)/(\mathrm dy)`. This is due to the symmetry of the original left hand side of the equation:


`f(y,x)=y^xx^y=x^yy^x=f(x,y)`