Answer:To find the derivative
dx
dy
for the function
=
sin
(
)
tan
(
)
y=sin(x)tan(x), we need to use the product rule for differentiation. The product rule states that if
=
(
)
(
)
y=u(x)v(x), then:
=
(
)
+
(
)
dx
dy
=u(x)
dx
dv
+v(x)
dx
du
Step-by-step explanation:Here, we let:
(
)
=
sin
(
)
u(x)=sin(x) and
(
)
=
tan
(
)
v(x)=tan(x).
First, we find the derivatives of
(
)
u(x) and
(
)
v(x):
=
cos
(
)
dx
du
=cos(x)
=
sec
2
(
)
dx
dv
=sec
2
(x)
Now, we apply the product rule:
=
sin
(
)
[
tan
(
)
]
+
tan
(
)
[
sin
(
)
]
dx
dy
=sin(x)
dx
d
[tan(x)]+tan(x)
dx
d
[sin(x)]
Substituting the derivatives we found:
=
sin
(
)
sec
2
(
)
+
tan
(
)
cos
(
)
dx
dy
=sin(x)sec
2
(x)+tan(x)cos(x)
So the derivative
dx
dy
for the function
=
sin
(
)
tan
(
)
y=sin(x)tan(x) is:
=
sin
(
)
sec
2
(
)
+
tan
(
)
cos
(
)
dx
dy
=sin(x)sec
2
(x)+tan(x)cos(x)