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1. If f(x)=sin(x^2), Find f'(x) and f'(1)
2. If d/dx(f(3x^3))=9x^2, Calculate f'(x).
3. Let f(x)=cos(6x+6), find f'(x)
4. Let f(x)=-2sec(2x), find f'(x)


    Looks like an exercise in the chain rule:

    f(x) = sin(x^2)
    f'(x) = cos(x^2) * 2x = 2x cos(x^2)
    f'(1) = 2cos(1)

    Let g(x) = f(3x^3)
    dg/dx = dg/df * df/dx
    9x^2 = dg/df * 9x^2
    1 = dg/df
    Looks like g(x) = x
    so, g(f) = f
    dg/dx = df/dx = 9x^2
    f' = -sin(6x+6)*6
    = -6sin(6x+6)
    f' = -2sec(2x)tan(2x)*2
    = -2sec(2x)tan(2x)

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