# 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

Staff Emeritus
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.