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Math Help - rate of change dy/dx

  1. #1
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    Post rate of change dy/dx

    Find the rate of change (dy/dx) for the prescribed value of x.

    y = (x^2 + 2) (x + square root of x) x=4
    y = (x^2 + 2)(x + x^-1/2)
    y = (x^2 + 2)(1/2square root x) + (x + 1/2square root of x)(2x)

    If that is correct thus far, how do I go about multipling this out?
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  2. #2
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    Quote Originally Posted by becky
    Find the rate of change (dy/dx) for the prescribed value of x.
    y = (x^2 + 2) (x + square root of x) x=4
    y = (x^2 + 2)(x + x^-1/2)
    y = (x^2 + 2)(1/2square root x) + (x + 1/2square root of x)(2x)
    If that is correct thus far, how do I go about multipling this out?
    Hello,

    you've got:

    y=(x^2+2)(x+x^{\frac{1}{2}})

    Obviously you don't know the product rule to calculate the derivative of y. So you have to multiply first:

    y=(x^2+2)(x+x^{\frac{1}{2}})=x^3+x^{\frac{5}{2}}+2  x+2x^{\frac{1}{2}}

    Now differentiate:

    \frac{dy}{dc}=3x^2+{5\over 2}x^{\frac{3}{2}}+2+ x^{-\frac{1}{2}}

    Plug in the value you know and you'll get:

    3(4)^2+{5\over 2}(4)^{\frac{3}{2}}+2+ (4)^{-\frac{1}{2}} =

    48+20+2+ \frac{1}{2}} = \frac{141}{2}

    Greetings

    EB
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