Answer:
Differentiation will give you the gradient for the tangent at any point, and you use the product rule whenever a function can be thought of as two functions multiplied together.
If
f
(
x
)
=
g
(
x
)
×
h
(
x
)
then
f
'
(
x
)
=
g
'
(
x
)
h
(
x
)
+
g
(
x
)
h
'
(
x
)
so if
y
=
x
×
sin
x
then
d
y
d
x
=
1
×
sin
x
+
x
×
cos
x
=
sin
x
+
x
cos
x
We know that
x
=
π
2
, so the gradient is
m
=
sin
(
π
2
)
+
π
2
cos
(
π
2
)
=
1
+
π
2
×
0
=
1
Therefore, we can say that
y
=
m
x
+
c
y
=
(
1
)
x
+
c
y
=
x
+
c
So all we really need to find now is the value for
c
, the
y
intercept. We do this by working out a point
(
x
,
y
)
on the graph. We are already given that
x
=
π
2
, so
y
=
x
sin
x
=
π
2
sin
(
π
2
)
=
π
2
×
1
=
π
2
∴
(
x
,
y
)
=
(
π
2
,
π
2
)
Now we substitute this into the equation we already have for the tangent,
y
=
x
+
c
,
(
x
,
y
)
=
(
π
2
,
π
2
)
π
2
=
π
2
+
c
c
=
π
2
−
π
2
=
0
∴
y
=
x
+
c
=
x
+
(
0
)
=
x
which means the tangent to the curve
y
=
x
sin
x
at
(
π
2
,
π
2
)
is simply
y
=
x
.
