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Math Help - Help with derivative

  1. #1
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    Help with derivative

    I need to find the derivative of f(x)=x+\sqrt x+3

    I have gotten this far: f'(x)=lim_{h\to0}\frac{f(x+h)-f(x)}{h}=lim_{h\to0}\frac{(x+h)+\sqrt{x+h} +3-\left(x+\sqrt x +3\right)}{h}=lim_{h\to0}\frac{h+\sqrt{x+h}-\sqrt x}{h}

    At this point I have tried some crazy stuff to eliminate the h on the bottom. The closest I have come is this: lim_{h\to0}\frac{h}{h}+\frac{\sqrt{x+h}}{h}-\frac{\sqrt x}{h} which can be broken down into, lim_{h\to0}\frac{h}{h}+\frac{\sqrt x}{h}+\frac{\sqrt h}{h}-\frac{\sqrt x}{h} which leaves this: lim_{h\to0}\frac{h}{h}+\frac{\sqrt h}{h}, right?

    If I haven't confused myself along the way, then I should be on track, and the answer seems tantilizingly close, but I get lim_{h\to0}1+\frac{\sqrt h}{h}= lim_{h\to0}1+\frac{1}{\sqrt h}

    What am I doing wrong?
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  2. #2
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    Re: Help with derivative

    The step that you did incorrectly was the \sqrt((x+h))/h not (\sqrt(x)/h-(\sqrt(h)/h))
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  3. #3
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    Re: Help with derivative

    Quote Originally Posted by rortiz81 View Post
    \cdots+\frac{\sqrt{x+h}}{h}+\cdots which can be broken down into, \cdots+\frac{\sqrt x}{h}+\frac{\sqrt h}{h}+\cdots
    Absolutely not. Do not do this.

    \sqrt{a + b} \neq \sqrt a + \sqrt b

    Quote Originally Posted by rortiz81 View Post
    I need to find the derivative of f(x)=x+\sqrt x+3
    Rationalizing the numerator can be a good strategy. Continuing your work from before you tried to split everything up:

    f'(x) = \lim_{h\to0}\frac{h + \sqrt{x + h} - \sqrt x}h

    =\lim_{h\to0}\left(1 + \frac{\sqrt{x+h}-\sqrt x}h\right)

    =\lim_{h\to0}\left(1 + \frac{\sqrt{x+h}-\sqrt x}h\cdot\frac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}\right)

    =\lim_{h\to0}\left[1 + \frac{(x + h) - x}{h\left(\sqrt{x+h} + \sqrt x\right)}\right]

    =\lim_{h\to0}\left[1 + \frac h{h\left(\sqrt{x+h} + \sqrt x\right)}\right]

    =\lim_{h\to0}\left(1 + \frac1{\sqrt{x+h} + \sqrt x}\right)

    =1+\frac1{2\sqrt{x}}
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  4. #4
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    Re: Help with derivative

    If you can prove/use the fact that \frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x) for differentiable functions f and g, then the derivative is equal to

    \frac{d}{dx} x + \frac{d}{dx} \sqrt{x} - \frac{d}{dx} 3, which is equal to

    1 + \frac{1}{2 \sqrt{x}}
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  5. #5
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    Re: Help with derivative

    Thank you all. Ugh. I really should have better algebra skills than this. BTW, this forum rocks.
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