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Math Help - Derivative using limit definition and line tangent to graph

  1. #1
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    Derivative using limit definition and line tangent to graph

    Hi!

    Apparently I made mistakes when solving this, could I be helped in spotting the errors please?
    Thanks in advance!

    a. Given f(x)=\sqrt{x-2}+1, find f^{\prime}(x) using the limit definition of the derivative.

    f^{\prime}(x) = \frac1h \cdot \frac{h+2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}

    = \frac{2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}

    = \frac{2}{2\sqrt{x-2}+2}

    = \frac{1}{\sqrt{x-2}+2}

    b. Find the equation of the line tangent to the graph f(x)=\sqrt{x-2}+1 at the point x = 6.

    y = f(x)(f^{\prime}(x)(x-x_0)

    f^{\prime}(x) = \frac{1}{2+2} = \frac14

    f(x) = \left(\sqrt{6-2}+1\; = 3\right)

    y = 3\cdot\frac14(x-6)
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  2. #2
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    Re: Derivative using limit definition and line tangent to graph

    Quote Originally Posted by Unreal View Post
    Hi!

    Apparently I made mistakes when solving this, could I be helped in spotting the errors please?
    Thanks in advance!

    a. Given f(x)=\sqrt{x-2}+1, find f^{\prime}(x) using the limit definition of the derivative.
    rather than try and spot your error I'm just going to go through how it should be done.

    \frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{f[x+h] - f[x]}{h}

    \frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{(\sqrt{x+h-2}+1)-(\sqrt{x-2}+1)}{h}

    \frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{\sqrt{x+h-2}-\sqrt{x-2}}{h}

    \frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{(x+h-2)-(x-2)}{h(\sqrt{x+h-2}+\sqrt{x-2})}

    \frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{h}{h(\sqrt{x+h-2}+\sqrt{x-2})}

    \frac{d}{dx}f[x]=\text{limit}_{h\rightarrow0}\frac{1}{(\sqrt{x+h-2}+\sqrt{x-2})}

    \frac{d}{dx}f[x]=\frac{1}{2\sqrt{x-2}}

    ok here's this for now so you can look at it. You actually made only one small mistake in keeping track of a negative sign.
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  3. #3
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    Re: Derivative using limit definition and line tangent to graph

    Quote Originally Posted by Unreal View Post
    Hi!

    Apparently I made mistakes when solving this, could I be helped in spotting the errors please?
    Thanks in advance!

    a. Given f(x)=\sqrt{x-2}+1, find f^{\prime}(x) using the limit definition of the derivative.

    f^{\prime}(x) = \frac1h \cdot \frac{h+2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}
    The "difference quotient" would be \frac{(\sqrt{x+h- 2}+ 1)- (\sqrt{x-2}+1)}{h}= \frac{\sqrt{x+h- 2}- \sqrt{x- 2}}{h}
    Now multiplying numerator and denominator by \sqrt{x+h- 2}+ \sqrt{x- 2} we have \frac{x+ h- 2- (x- 2)}{h(\sqrt{x+h- 2}+ \sqrt{x- 2})}= \frac{h}{h(\sqrt{x+h- 2)+ \sqrt{x- 2})}. That is, there is no "+ 2" in the numerator.

    = \frac{2}{\left(\sqrt{x+h-2}+1+\sqrt{x-2}+1\right)}

    = \frac{2}{2\sqrt{x-2}+2}

    = \frac{1}{\sqrt{x-2}+2}

    b. Find the equation of the line tangent to the graph f(x)=\sqrt{x-2}+1 at the point x = 6.

    y = f(x)(f^{\prime}(x)(x-x_0)

    f^{\prime}(x) = \frac{1}{2+2} = \frac14

    f(x) = \left(\sqrt{6-2}+1\; = 3\right)

    y = 3\cdot\frac14(x-6)
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