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patbatE21

  • 2 years ago

Find f' of f(x)=sin(2x)^2 at f'(pi/6)

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  1. .Sam.
    • 2 years ago
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    f(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))

  2. .Sam.
    • 2 years ago
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    f'(pi/6)=4sin((2[pi/6]))cos((2[pi/6]))

  3. patbatE21
    • 2 years ago
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    COuld you look at this Prove that 76^(130) + 1 is not prime

  4. .Sam.
    • 2 years ago
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    I got to go now , you can tell others to solve that

  5. patbatE21
    • 2 years ago
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    ok thanks though

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