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Science
:
Math Exam Resources/Courses/MATH100/December 2011/Question 01 (k)/Solution 1
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From UBC Wiki
<
Science:Math Exam Resources
|
Courses/MATH100
|
December 2011
|
Question 01 (k)
We compute this derivative using the chain rule twice and the product rule
d
y
d
t
=
e
t
cos
(
2
t
)
⋅
d
d
t
(
t
cos
(
2
t
)
)
=
e
t
cos
(
2
t
)
⋅
(
d
d
t
(
t
)
⋅
cos
(
2
t
)
+
t
d
d
t
(
cos
(
2
t
)
)
)
=
e
t
cos
(
2
t
)
⋅
(
1
⋅
cos
(
2
t
)
+
t
(
−
sin
(
2
t
)
⋅
2
)
)
=
e
t
cos
(
2
t
)
(
cos
(
2
t
)
−
2
t
sin
(
2
t
)
)
{\displaystyle {\begin{aligned}{\frac {dy}{dt}}&=e^{t\cos(2t)}\cdot {\frac {d}{dt}}(t\cos(2t))\\&=e^{t\cos(2t)}\cdot ({\frac {d}{dt}}(t)\cdot \cos(2t)+t{\frac {d}{dt}}(\cos(2t)))\\&=e^{t\cos(2t)}\cdot (1\cdot \cos(2t)+t(-\sin(2t)\cdot 2))\\&=e^{t\cos(2t)}\left(\cos(2t)-2t\sin(2t)\right)\end{aligned}}}