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Math Help - equation of the tangent line to the curve

  1. #1
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    equation of the tangent line to the curve

    determine the equation of the tangent line to the curve
    f(x)=4x^3(x-2)^4(6x-5)^5 at point (1,4)

    can u just help to get F'(x) then i know how to plug 1 in f'(X).

    please help and show solution
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Max10 View Post
    determine the equation of the tangent line to the curve
    f(x)=4x^3(x-2)^4(6x-5)^5 at point (1,4)

    can u just help to get F'(x) then i know how to plug 1 in f'(X).

    please help and show solution
    use the product rule (with the chain rule where necessary) and note that:

    if f, g, \text{ and } h are functions of x, then

    \frac d{dx} [fgh] = f'gh + fg'h + fgh'

    here you can let f = 4x^3, g = (x - 2)^4, and h = (6x - 5)^5

    and note that you need the chain rule to find g' and h'
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  3. #3
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    so is it f'(x) 12x^2 g'(x)=4(x-2)^3(x) h'(x)=5(6x-5)^4(6)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Max10 View Post
    so is it f'(x) 12x^2 g'(x)=4(x-2)^3(x) h'(x)=5(6x-5)^4(6)
    your g'(x) is wrong
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    your g'(x) is wrong

    how come (x-2)^4

    g'(x) 4(x-2)^3(x)

    please help
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Max10 View Post
    how come (x-2)^4

    g'(x) 4(x-2)^3(x)

    please help
    the derivative of x - 2 is not x...
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