# derivative of sin^2

by elsternj
Tags: derivative, sin2
 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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P: 9,675
 Quote by elsternj 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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P: 38,706
 Quote by 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?
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!

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.

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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