Taking
logarithms, we have
log y = log (sin x)sin x = sin x log (sin x)
Then
1 dy
y dx  = d
dx  (sin x log (sin x))
= cos x log (sin x) + sin x 1
(sin )
sin
d
x
x dx
⋅
= cos x log (sin x) + cos x
= (1 + log (sin x)) cos x
Thus
dy
dx  = y((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x
Example 42 For a positive constant a find dy
dx , where
1
1
, and
a
t t
y
a
x
t
t
+


=
=
+



Solution Observe that both y and x are defined for all real t ≠ 0 Clearly
dy
dt  = 
(
)
t t1
d
a
dt
+
 =
1
1
log
at t d
t
a
dt
t
+

+
⋅




=
1
12
1
log
at t
a
t
+ 

−



Similarly
dx
dt  =
1
1
1
a
d
a t
t
t
dt
t
−




+
⋅
+







=
1
2
1
1
1
a
a t
t
t
−




+
⋅
−







dx
dt  ≠ 0 only if t ≠ ± 1 Thus for t ≠ ± 1,
Rationalised 2023-24
 MATHEMATICS
144
dy
dy
dt
dx
dx
dt
=
 =
a
t
a
a t
t
t
t t
a
+
−
−


+

⋅
−


1
2
1
2
1
1
1
1
1
log
=
1
1
log
1
t t
a
a
a
a t
t
+
−

+



Example 43 Differentiate sin2 x w