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Mathematics
Mathematics
Problem
Let
x
=
e
2
t
x=e^{2t}
x
=
e
2
t
and
y
=
e
t
+
t
y=e^{t}+t
y
=
e
t
+
t
. Determine
d
y
d
x
\dfrac{dy}{dx}
d
x
d
y
at the value of
t
t
t
that satisfies
sin
t
=
1
2
\sin t=\dfrac{1}{2}
sin
t
=
2
1
with
0
<
t
<
π
0<t<\pi
0
<
t
<
π
. a)
e
π
/
6
+
1
e
π
/
3
\dfrac{e^{\pi/6}+1}{e^{\pi/3}}
e
π
/3
e
π
/6
+
1
b)
e
π
/
6
+
1
2
e
π
/
3
\dfrac{e^{\pi/6}+1}{2e^{\pi/3}}
2
e
π
/3
e
π
/6
+
1
c)
2
e
π
/
3
e
π
/
6
+
1
\dfrac{2e^{\pi/3}}{e^{\pi/6}+1}
e
π
/6
+
1
2
e
π
/3
d)
e
π
/
6
+
1
2
e
π
/
6
\dfrac{e^{\pi/6}+1}{2e^{\pi/6}}
2
e
π
/6
e
π
/6
+
1
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