Find the derivative of y = 2cos(-4x) using chain rule

[K_Swiss]

y = 2cos(-4x)

y' = 2cos(-4x)-sinx
y' = -2cos(-4x)sinx

Textbook Answer: y' = 8sin(-4x) ---- please explain how to get this answer???

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y = cos?x

y' = cosx^(0.5)
y' = 0.5(cosx)(-sinx)
y' = -0.5(cosx)^(-0.5)(sinx)
y' = -0.5sinx / cosx^(0.5)
y' = -0.5sinx / cos?x

Textbook Answer: y' = - 1 / 2?x ---- please explain how to get this answer???

 

[stapel]

y = 2cos(-4x)

y' = 2cos(-4x)-sinx

How did you differentiate the cosine to get a product of a cosine and a sine? Shouldn't the derivative of the cosine be just the sine? And where did you continue on (using the Chain Rule) to differentiate the "inner" function, "-4x"?

y = cos?x

y' = cosx^(0.5)
y' = 0.5(cosx)(-sinx)

Review the derivative of f(x) = cos(x). It is not cos(x)sin(x).

Also, review the Power Rule: the derivative of g(x) = x[sup:e19awmok]1/2[/sup:e19awmok] is not (1/2)x.

Eliz.