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Math Help - find derivative of sqrtx...

  1. #1
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    find derivative of sqrtx...

    Use the definition to find the derivative of f(x)= \sqrt{x}, for >0. Is f differential at zero? Explain.

    I believe the def we need is let f: D->R with x_0 an accumulation point of D and x_0 \in D.with x not equal to x_0, define T(x)=f(x)-f(x_0)/x-x_0. The function f is said to be differentiable at x_0 iff T has a limit at x_0 and we write lim_x->x_0 T(x)=f'(x_0). ..I understand it I'm just sure how to show a proof and stuff..
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  2. #2
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    Derivative

    Hello,

    If you would like to find the derivative of the function f(x)= \sqrt{x}, then perhaps you could follow the 3-step rule:

    Let f be a function and x_{0} a point in the domain, then do as follows:

    1) Compute \triangle y=f(x_{0}+h)-f(x_{0})

    2) Compute and reduce \frac{\triangle y}{h}=\frac{f(x_{0}+h)-f(x_{0})}{h}

    3) Let h\to \infty in the reduced result from 2). If this limit exists then this limit will be the derivative, and the function f differentiable.
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  3. #3
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    Quote Originally Posted by surjective View Post
    Hello,

    If you would like to find the derivative of the function f(x)= \sqrt{x}, then perhaps you could follow the 3-step rule:

    Let f be a function and x_{0} a point in the domain, then do as follows:

    1) Compute \triangle y=f(x_{0}+h)-f(x_{0})

    2) Compute and reduce \frac{\triangle y}{h}=\frac{f(x_{0}+h)-f(x_{0})}{h}

    3) Let h\to \infty in the reduced result from 2). If this limit exists then this limit will be the derivative, and the function f differentiable.
    isnt that basically the same def i posted?
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  4. #4
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    Quote Originally Posted by surjective View Post
    Hello,

    If you would like to find the derivative of the function f(x)= \sqrt{x}, then perhaps you could follow the 3-step rule:

    Let f be a function and x_{0} a point in the domain, then do as follows:

    1) Compute \triangle y=f(x_{0}+h)-f(x_{0})

    2) Compute and reduce \frac{\triangle y}{h}=\frac{f(x_{0}+h)-f(x_{0})}{h}

    3) Let h\to \infty in the reduced result from 2). If this limit exists then this limit will be the derivative, and the function f differentiable.
    would that give me \sqrt{h}/h ?
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  5. #5
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    Quote Originally Posted by alice8675309 View Post
    would that give me \sqrt{h}/h ?
    Firstly, yes your definition is correct. You need to let h \to 0 instead of \infty, though..

    Your result is not correct - \sqrt{x+h} - \sqrt{x} \neq \sqrt{x+h-x}... Instead, multiply by conjugate:

    \frac{\sqrt{x+h} - \sqrt{x}}{h} = \frac{(\sqrt{x+h} - \sqrt{x})}{h} \cdot \frac{(\sqrt{x+h} + \sqrt{x})}{ (\sqrt{x+h} + \sqrt{x})}

    And from there, simplify and then let h \to 0 to get the result.
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  6. #6
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    Quote Originally Posted by Defunkt View Post
    Firstly, yes your definition is correct. You need to let h \to 0 instead of \infty, though..

    Your result is not correct - \sqrt{x+h} - \sqrt{x} \neq \sqrt{x+h-x}... Instead, multiply by conjugate:

    \frac{\sqrt{x+h} - \sqrt{x}}{h} = \frac{(\sqrt{x+h} - \sqrt{x})}{h} \cdot \frac{(\sqrt{x+h} + \sqrt{x})}{ (\sqrt{x+h} + \sqrt{x})}

    And from there, simplify and then let h \to 0 to get the result.
    ok so now i have lim_h->0 1/ \sqrt{x+h}+ \sqrt{x} and then as the limit of h goes to 0 we have 1/2 \sqrt{x}, which means yes f is differentiable at zero, b/c as h goes to zero the derivative equals 1/2 \sqrt{x}
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  7. #7
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    Also, to put a fraction in math tags:

    \frac{\sqrt{x+h} - \sqrt{x}}{h} produces \frac{\sqrt{x+h} - \sqrt{x}}{h}
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