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!
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
Re: Derivative of Sin^2(2x)? You first let U= sin 2x so that [itex]y= U^2[/itex], [itex]y'= 2U U'[/itex]. 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 [itex]sin^2(x)[/itex] is not [itex]cos^2(x)[/itex]
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.