Solving the Derivative of Sin22x

[elsternj]

Homework Statement

Derivative of Sin²2x

Homework Equations

dy/dx = dy/du * du/dx

y=U²

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!

 

[ehild]

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.

ehild

 

[HallsofIvy]

Homework Statement

Derivative of Sin²2x

Homework Equations

dy/dx = dy/du * du/dx

y=U²

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 y= U^2, y'= 2U U'.

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 sin^2(x) is not cos^2(x)

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

 

[nickalh]

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.