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anonymous
 4 years ago
Find f' of f(x)=sin(2x)^2 at f'(pi/6)
anonymous
 4 years ago
Find f' of f(x)=sin(2x)^2 at f'(pi/6)

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.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1f(x)=sin(2x)^(2) The derivative of f(x) is equal to f'(x)=(d)/(dx) sin(2x)^(2). f'(x)=(d)/(dx) sin(2x)^(2) Remove the parentheses around the expression cos(v). (d)/(dv) sin(v)=cos(v) Simplify the derivative. (d)/(dx) sin^(2)(2x)=4sin((2x))cos((2x)) The derivative of the function is 4sin((2x))cos((2x)). f'(x)=4sin((2x))cos((2x))

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1f'(pi/6)=4sin((2[pi/6]))cos((2[pi/6]))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0COuld you look at this Prove that 76^(130) + 1 is not prime

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.1I got to go now , you can tell others to solve that
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