Given x+|y|=2y ⇒x+y=2y or x−y=2y ⇒x=y or x=3y This represent a straight line which passes through origin. Hence, x+|y|=2y is continuous at x=0. Now, we check differentiability at x=0 x+|y|=2y⇒x+y=2y,y≥0 x−y=2y,y<0 Thus, f(x)={
x,
y<0
x∕3,
y≥0
} Now, L.H.D.=
lim
h→0−
f(x+h)−f(x)
−h
=
lim
h→0−
x+h−x
−h
=−1 R.H.D=
lim
h→0+
f(x+h)−f(x)
h
=
lim
h→0+
x+h
3
−
x
3
h
=
lim
h→0+
1
3
=
1
3
Since, L.H.D ≠ R.H.D. at x = 0 ∴ given function is not differentiable at x = 0