Question

if e^y=y^x then prove that
dy/dx =(log y)²/(log y-1)​

Answer 1

We have , yx=ey−x

Taking log both side

⇒logyx=logey−x

⇒xlogy=y−x....(1)

Differentiating w.r.t x, we get

1.logy+xy1dxdy=dxdy−1

⇒dxdy=1−yxlogy+1

             =y−1+logyyy(logy+1)   [Using (1)]

             =logy(1+logy)2