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Math Help - first princple

  1. #1
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    first princple

    Hi,can someone help me in finding the limit for (5x)^3/2 using first principle.
    ((5x+h)^3/2-(5x)^3/2)/h
    This is all i got and then I am stuck on it.
    Help would be greatly appreciated.
    Thanx
    One more
    (3x+5)^3/2
    Thanx in advance
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  3. #3
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    Hello, crystal91!

    This is a killer!


    Differentiate f(x)\,=\,(5x)^{\frac{3}{2}} using first principles.
    Note that: . f(x) \;=\;(5x)^{\frac{3}{2}} \;=\;5^{\frac{3}{2}}\cdot x^{\frac{3}{2}} \;=\; 5\sqrt{5}\,x^{\frac{3}{2}}

    We will differentiate f(x) \,=\,x^{\frac{3}{2}} . . . and multiply by 5\sqrt{5} at the very end.


    [1] Find f(x+h) - f(x)

    . . f(x+h) - f(x) \;\;=\;\;(x+h)^{\frac{3}{2}} - x^{\frac{3}{2}} \;\;=\;\;\left(\sqrt{x+h}\right)^3 - \left(\sqrt{x}\right)^3 . . . a difference of cubes

    . . Factor:. . \left(\sqrt{x+h} - \sqrt{x}\right)\,\left(x+h + \sqrt{x}\,\sqrt{x+h} + x\right) . =\;\;\left(\sqrt{x+h} - \sqrt{x}\right)\,\left(2x + h + \sqrt{x(x+h)}\right)

    . . Multiply by \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\!:\quad \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}\cdot\frac{(\sqrt{x+h} - \sqrt{x})\,(2x + h + \sqrt{x(x+h)}}{1}

    . . and we have: . \frac{[(x+h)-x]\,[2x + h + \sqrt{x(x+h)}]}{\sqrt{x+h} + \sqrt{x}}  \;=\;\frac{h\,(2x+h + \sqrt{x(x+h)})}{\sqrt{x+h} + \sqrt{x}}


    [2] Divide by h\!:\quad \frac{2x+h + \sqrt{x(x+h)}}{\sqrt{x+h} + \sqrt{x}}


    [3] Take the limit: . \lim_{h\to0}\left[\frac{2x+h+\sqrt{x(x+h)}}{\sqrt{x+h} + \sqrt{x}}\right] \;=\;\frac{2x+0+\sqrt{x(x+0)}}{\sqrt{x+0} + \sqrt{x}}

    . . . . . . . . . . . . =\;\frac{2x + \sqrt{x^2}}{\sqrt{x} + \sqrt{x}} \;\;=\;\;\frac{2x + x}{2\sqrt{x}} \;\;=\;\;\frac{3x}{2\sqrt{x} }\;\;=\;\;\frac{3}{2}\sqrt{x}


    Multiply by 5\sqrt{5}\!:\;\;f'(x) \;=\;\frac{15\sqrt{5}}{2}\,\sqrt{x}


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