Taking logarithm on both sides, we have
log u = x log y
Differentiating both sides w r t x, we have
1 du
u dx
⋅
 =
(log )
log
( )
d
d
x
y
y
x
dx
dx
+
=
1
log
1
dy
x
y
y dx
⋅
+
⋅
So
du
dx  =
log
log
x
x dy
x dy
u
y
y
y
y dx
y dx




+
=
+







