Question

Differenciate, Y = Sin (1 - 2x)³ .​

Answer 1

Given, y=(sin(1-2x)^3

We will use chain rule to differenitaiate it

dy/dx=dy/dx {sin(1-2x)^3} ×dy/dx sin(1-2x)×dy/dx(1-2x)

=>3sin^2(1-2x)×cos(1-2x)×-2

=>-6sin^2(1-2x)cos(1-2x)

simplifying more we get

=>-3×sin(1-2x)×2sin(1-2x)cos(1-2x)

=>-3sin(1-2x)×sin2(1-2x) is the answer.

Cheers~!!

Answer 2

d
y
d
x
=
1
√
−
x
(
x
+
1
)
.
Explanation:
It is known that
d
d
t
sin
−
1
t
=
1
√
1
−
t
2
.
Let
(
2
x
+
1
)
=
t
;
.
y
=
sin
−
1
t
,
t
=
2
x
+
1
.
Thus,
y
is a function of
t
, and,
t
of
x
.
Therefore, by Chain Rule,
d
y
d
x
=
d
y
d
t
⋅
d
t
d
x
...
...
...
...
...
...
(
⋆
)
Now,
y
=
sin
−
1
t
⇒
d
y
d
t
=
1
√
1
−
t
2
...
...
...
...
...
.
(
1
)
t
=
2
x
+
1
⇒
d
t
d
x
=
2
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
.
.
(
2
)
Using
(
1
)
,
(
2
)
in
(
⋆
)
, &, remembering that
t
=
2
x
+
1
, we get,
d
y
d
x
=
(
1
√
1
−
t
2
)
(
2
)
=
2
√
1
−
(
2
x
+
1
)
2
=
2
√
1
−
4
x
2
−
4
x
−
1
Therefore,
d
y
d
x
=
1
√
−
x
(
x
+
1
)
.