By putting 2x = tan θ, this function may be
rewritten as
f(x) =
1
1
2
sin
1
4
x
x
+
− 



+

= sin−
⋅
+ (
)






1
2
2
2
1
2
x
x
=
1
2tan2
sin
1
tan
−
θ




+
θ


= sin –1 [sin 2θ]
= 2θ = 2 tan – 1 (2x)
Thus
f ′(x) =
(
)
2
1
2
(2 )
1
2
x
x
dxd
⋅
⋅
+
=
2
(2 )log2
1
4
x
x ⋅
+
=
1
2
log2
1
4
x
x
+
+
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
143
Example 41 Find  f ′(x) if f(x) = (sin x)sin x for all 0 < x < π Solution The function y = (sin x)sin x is defined for all positive real numbers 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