log(√x2+y2)
Letx2+y2=t 2x+dx+2ydy=dt
2x+2ydydx=dtdx
ddxlog√t(ddtlog√t)(dtdx)=(1√t.12√t)(2x+2yy′)=12t/2(x+yy′)1=1x2y2(x+yy′) ......(2)
ddxtan−1(xy)
Let xy=tx=yt
or dx=dy+ydt
1−dydx+ydtdxdtdx=1−dydxtyddxtan−1=ddt(tan−1t)dtdx=11+t2(1−y′ty)=11+t2(1−y′ty)=11+x2y2{1−y′(xy)y}=y2y2+x2{y−y′(x)y2} ---(1)
Equality (1)&(2)
y−y′xx2+y2=x+yy′x2+y2y−x=y′x+y′y=y′(x+y)∴y′=y−xx+y