Yahoo Answers: Answers and Comments for Help with finding stationary point of y=e^x•sin(2x)? [Mathematics]
Copyright © Yahoo! Inc. All rights reserved.
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
From Diana
enUS
Tue, 25 Jun 2019 14:16:07 +0000
3
Yahoo Answers: Answers and Comments for Help with finding stationary point of y=e^x•sin(2x)? [Mathematics]
292
38
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
https://s.yimg.com/zz/combo?images/emaillogous.png

From Some Body: Stationary points are where the derivative is ...
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
Tue, 25 Jun 2019 14:38:29 +0000
Stationary points are where the derivative is 0.
Take the derivative using product rule and chain rule.
y' = eˣ cos(2x) (2) + eˣ sin(2x)
y' = eˣ (2 cos(2x) + sin(2x))
Set equal to 0.
0 = eˣ (2 cos(2x) + sin(2x))
eˣ is never 0, so:
0 = 2 cos(2x) + sin(2x)
0 = 2 + tan(2x)
tan(2x) = 2
2x = atan(2) + kπ
x = ½ atan(2) + k/2 π
Trying different integer values of k:
k = 0, x ≈ 0.554
k = 1, x ≈ 1.017
k = 2, x ≈ 2.588
Only k = 1 gives us an answer within 0≤x≤0.5π.
Therefore, the stationary point is at x = ½ atan(2) + π/2, or:
(x, y) ≈ (1.017, 2.474)

From alex: Hint:
f '(x) = 0 <> Stationary p...
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
Wed, 26 Jun 2019 01:18:07 +0000
Hint:
f '(x) = 0 <> Stationary points

From la console: y = e^(x) * sin(2x) ← this is a function i.e. ...
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
Tue, 25 Jun 2019 14:55:44 +0000
y = e^(x) * sin(2x) ← this is a function i.e. a curve
You can obtain the stationary point when the derivative is zero.
The function looks like (u.v), so the derivative looks like: (u'.v) + (v'.u) → where:
u = e^(x) → u' = e^(x)
v = sin(2x) → v' = 2.cos(2x)
y' = (u'.v) + (v'.u)
y' = [e^(x) * sin(2x)] + [2.cos(2x) * e^(x)]
y' = e^(x) * [sin(2x) + 2.cos(2x)] → then you solve for x the equation: y' = 0
e^(x) * [sin(2x) + 2.cos(2x)] = 0 → you know that: e^(x) ≠ 0 of course
sin(2x) + 2.cos(2x) = 0
sin(2x) =  2.cos(2x)
sin(2x)/cos(2x) =  2
tan(2x) =  2 ← the corresponding angle is ≈  1.107148 rd → i.e.: 2.034444 rd
2x = 2.034444 + kπ → where k is an integer
x = 1.017222 + k.(π/2)
First case: k = 0 → x = 1.017222 ← this is the solution where: 0 ≤ x ≤ 0.5π
Second case: k = 1 → x = 1.017222 + (π/2) ← over 0.5π
Third case: k = 2 → x = 1.017222 + π ← over 0.5π

From billrussell42: In mathematics, particularly in calculus, a st...
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
https://answers.yahoo.com/question/index?qid=20190625141607AAit0Mj
Tue, 25 Jun 2019 14:47:22 +0000
In mathematics, particularly in calculus, a stationary point of a differentiable function of one variable is a point on the graph of the function where the function's derivative is zero.
y = e^x•sin(2x)
I assume that that is
y = (e^x)•sin(2x)
y' = (e^x)•sin(2x) + (e^x)•2cos(2x) = 0
sin(2x) + 2cos(2x) = 0
2cos(2x) = cos(2x+(π/2))
solve for x
x = about π/3, a bit less, approx 0.343π or 1.076
product rule
(fg)' = f'g + fg'
(fgu)' = fgu' + fug' + guf'
(e^x)' = e^x
(sin ax)' = a cos ax