xy−yx=ab
Putting u=xy and v=yx, we get
u−v=ab
Differentiating with respect to x, we get
dudx−dvdx=0⋯(1)
Now, u=xy
Taking logarithm on both sides, we get
logu=ylogx
Differentiating with respect to x, we get
1ududx=y×1x+logx×dydx
⇒dudx=u[yx+logx×dydx]
⇒dudx=xy[yx+logx×dydx]
Also, v=yx
Taking logarithm on both sides, we get
logv=xlogy
Differentiating with respect to x, we get
1vdvdx=x×1ydydx+logy×1
⇒dvdx=v[xydydx+logy]
⇒dvdx=yx[xydydx+logy]
From (1), we get
xy[yx+logx×dydx]−yx[xydydx+logy]=0
⇒[xylogx−xyx−1]dydx=yxlogy−y×xy−1
⇒dydx=yxlogy−y×xy−1xylogx−xyx−1