Question

If $x^{y} = a^{x}$ then prove that  dy/dx = xloga - y/ x logx

Answer 1

taking ㏒ both sides,
y㏒x = x㏒a

now differentiating both side with respect to x, 
dy/dx(㏒x) +y/x = ㏒a

⇒dy/dx = (㏒a -y/x)/㏒x
            =(x㏒a -y)/x㏒x