
#1
Jul2111, 03:41 PM

P: 42

1. The problem statement, all variables and given/known data
Derivative of Sin^{2}2x 2. Relevant equations dy/dx = dy/du * du/dx y=U^{2} 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=U^{2} then y` = 2Sin2x * cos2x? Or if U = 2x. y = SinU^{2} y` = 2cos2x * 2 y` = 4cos2x am i on the right track with either of these? any help is appreciated! thanks! 



#2
Jul2111, 04:11 PM

HW Helper
Thanks
P: 9,818

Sin^2(2x)=F(U(V)): V=2x, U=sin(V), F=U^2. dF/dx=dF/dU*dU/dV*dV/dx. ehild 



#3
Jul2111, 07:36 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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



#4
Jul2211, 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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