# Homework Help: Derivative of sin^2

1. Jul 21, 2011

### elsternj

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!

Last edited by a moderator: Feb 14, 2013
2. Jul 21, 2011

### ehild

Re: Derivative of Sin^2(2x)?

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

3. Jul 21, 2011

### HallsofIvy

Re: Derivative of Sin^2(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.

No, because the derivative of sin2X is not cos2X. Use the chain rule again.

No, because the derivative of $sin^2(x)$ is not $cos^2(x)$

4. Jul 22, 2011

### nickalh

Re: Derivative of Sin^2(2x)?

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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