derivative of sin^2


by elsternj
Tags: derivative, sin2
elsternj
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#1
Jul21-11, 03:41 PM
P: 42
1. The problem statement, all variables and given/known data
Derivative of Sin22x



2. Relevant equations
dy/dx = dy/du * du/dx

y=U2


3. The attempt at a solution
Just want to make sure I am doing this right*.

Do I let U = Sin2x or U = 2x?

Let's say U = Sin2x
y=U2

then y` = 2Sin2x * cos2x?

Or if U = 2x.

y = SinU2

y` = 2cos2x * 2
y` = 4cos2x

am i on the right track with either of these? any help is appreciated! thanks!
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ehild
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#2
Jul21-11, 04:11 PM
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Quote Quote by elsternj View Post
am i on the right track with either of these?
You are halfway on the right track .

Sin^2(2x)=F(U(V)): V=2x, U=sin(V), F=U^2.

dF/dx=dF/dU*dU/dV*dV/dx.

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HallsofIvy
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#3
Jul21-11, 07:36 PM
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Quote Quote by elsternj View Post
1. The problem statement, all variables and given/known data
Derivative of Sin22x



2. Relevant equations
dy/dx = dy/du * du/dx

y=U2


3. The attempt at a solution
Just want to make sure I am doing this right*.

Do I let U = Sin2x or U = 2x?
You first let U= sin 2x so that [itex]y= U^2[/itex], [itex]y'= 2U U'[/itex].

Then, to find U', let V= 2x so U= sin V. U'= cos(V)(V') and, of course, V'= 2.
Put those together.

Let's say U = Sin2x
y=U2

then y` = 2Sin2x * cos2x?
No, because the derivative of sin2X is not cos2X. Use the chain rule again.

Or if U = 2x.

y = SinU2

y` = 2cos2x * 2
y` = 4cos2x
No, because the derivative of [itex]sin^2(x)[/itex] is not [itex]cos^2(x)[/itex]

am i on the right track with either of these? any help is appreciated! thanks!

nickalh
nickalh is offline
#4
Jul22-11, 12:40 AM
P: 72

derivative of sin^2


Essentially, what this entire question boils down to:
We need two applications of the chain rule.
The first one started well.

Instead of
y` = 2Sin2x * cos2x
I recommend beginning Calculus students write.
y` = 2Sin2x * ( Sin2x )'
The left factor, 2 Sin(2x) is finished.
Then to evaluate the derivative of Sin 2x, apply chain rule a second time, with v =2x



General hint, way to think of chain rule:
Take deriv. of the outside, leave the inside alone , then multiply by deriv. of inside.


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