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Maths

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Differentiate y=(5x-2)^3(3-x)^6 using the product rule.

i don't get the step where they find the common factor could you help me by doing it step by step please ?
thank you!

  • Maths -

    There is no such step that i have ever heard of.

    Let
    u (x) = (5x-2)^3 and
    v (x) = (3-x)^6

    You want the derivative of u*v with respect to x.

    dy/dx = d(uv)/dx = u dv/dx + v du/dx
    That is the product rule.

    Now all you have to calculate is du/dx and dv/dx, multiply the v and u terms and add the result.

  • Maths -

    thanks!
    there is an example here and where i get lost is a step where it goes from
    dy/dx = d(uv)/dx = u dv/dx + v du/dx
    = (5x-2)^3[-6(3-x)^5]+(3-x)^6[15(5x-2)^2]
    = (5x-2)^2(3-x)^5(-30x+12+45-15x)

    is there a step between there that you could show me how to do so i can understand it ?

    thank you!

  • Maths -

    You must be talking about the simplication after you found the derivative.

    look at the line ....

    (5x-2)^3[-6(3-x)^5]+(3-x)^6[15(5x-2)^2]

    = -6(5x-2)^3(3-x)^5 + 15(3-x)^6(5x-2)^2


    now I see the following common factors
    -3 for the constants
    (5x-2)^2 and
    (3-x)^5

    so

    = -3(5x-2)^2(3-x)^5[2(5x-2) - 5(3-x)]
    = -3(5x-2)^2(3-x)^5[10x - 4 - 15 + 5x]
    = -3(5x-2)^2(3-x)^5(15x - 19)

    Your last line is not fully factored.

  • Maths -

    ah OK, now i get it.
    Thank you so much!!

  • Maths -

    hey sorry, one more thing.
    How do you find -3

    as in
    now I see the following common factors
    -3
    how ?

  • Maths -

    (-30x+12+45-15x)
    =-3(15x-19)

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