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

$\dfrac{\mathrm d}{\mathrm dx}[y\ln x+x\ln y]=0$
$\dfrac{\mathrm dy}{\mathrm dx}\ln x+\dfrac yx+\ln y+\dfrac xy\dfrac{\mathrm dy}{\mathrm dx}=0$
$\dfrac{\mathrm dy}{\mathrm dx}\left(\ln x+\dfrac xy\right)=-\dfrac yx-\ln y$
$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{\frac yx+\ln y}{\frac xy+\ln x}$
$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{y^2+xy\ln y}{x^2+xy\ln x}$

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 $\dfrac{\mathrm dx}{\mathrm dy}$. This is due to the symmetry of the original left hand side of the equation:

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