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P3 differentiation (again) watch

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    review exercise 1 Q55

    b)Given that y=xsin3x, prove that f"(x) + 9f(x) = 6cos3x

    f(x) = xsin3x
    f'(x) = x3cos3x +sin3x
    f"(x) = -x9sin3x + 3cos3x

    f"(x) + 9f(x) = 3cos3x - 9xsin3x +9xsin3x = 3cos3x not 6cos3x

    Where have I gone wrong. Prob some stupid mistake, I'll rep the person who spots it first.

    MB
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    (Original post by musicboy)
    review exercise 1 Q55

    b)Given that y=xsin3x, prove that f"(x) + 9f(x) = 6cos3x

    f(x) = xsin3x
    f'(x) = x3cos3x +sin3x

    f'(x) is 3cos3x + sin3x

    Edit: sorry.....forget that
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    (Original post by musicboy)
    review exercise 1 Q55

    b)Given that y=xsin3x, prove that f"(x) + 9f(x) = 6cos3x

    f(x) = xsin3x
    f'(x) = x3cos3x +sin3x
    f"(x) = -x9sin3x + 3cos3x

    f"(x) + 9f(x) = 3cos3x - 9xsin3x +9xsin3x = 3cos3x not 6cos3x

    Where have I gone wrong. Prob some stupid mistake, I'll rep the person who spots it first.

    MB
    f(x)= xsin3x
    f'(x)=3xcos3x+sin3x
    f''(x)=-9xsin3x+3cos3x+3cos3x
    =-9xsin3x+6cos3x

    So f''(x)+9f(x)=-9xsin3x+6cos3x+9xsin3x
    =6cos3x

    Hope that helps.
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    f''(x) = -9x*sin3x + 3cos3x + 3cos3x
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    Thanks comrades, I'll rep you both but Bezza will have to wait a day for his (sorry)

    MB
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    (Original post by musicboy)
    Thanks comrades, I'll rep you both but Bezza will have to wait a day for his (sorry)

    MB
    Thanks, but there's no need really. Happy to help.
 
 
 
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