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Equation of tangent line at sqrt(x)

  1. Oct 14, 2006 #1
    Hey, right now in our calculus course we are currently working on "derivatives" using the differentiation equations. I was hoping that someone could help me out with the following question:

    Find the equation of the tangent line to the curve at the given point.

    y = rootX , (4, 0.4)
    X + 1

    It's not an overly difficult question because we know that the equation is y-0.4 = m(x-4) and we are looking for m(the slope), and we also know that we should be using the formula f'(x)g(x)-g(x)f'(x)/[g(x)]^2

    However, I'm still having some difficulties in working this out, if someone could please shed some light on this, I'd be happy. :D
  2. jcsd
  3. Oct 14, 2006 #2
    So [tex] y = \frac{\sqrt{x}}{x+1} [/tex]. Have you used the quotient rule?

    [tex] \frac{dy}{dx} = \frac{(x+1)\frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}}{(x+1)^{2}} [/tex]

    Can you simplify this? You could just put in x = 4, without simplifying. Do you understand how to get the above expression?

    [tex] f(x) = \sqrt{x}[/tex] and [tex] g(x) = x+1 [/tex]

    So what is [tex] f'(x) [/tex] and [tex] g'(x) [/tex]?
    Last edited: Oct 14, 2006
  4. Oct 14, 2006 #3


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    Well, you have the method, so now work it out step by step:

    [tex] f(x) = \sqrt{x} [/tex]

    [tex] g(x) = x+1 [/tex]

    [tex] \left(\frac{f(x)}{g(x)}\right)' = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} [/tex]

    Anyway, this is your cue to take over:

    [tex] f'(x) = \textrm{?} [/tex]

    [tex] g'(x) = \textrm{?} [/tex]

    Substitute these expressions into the expression for the derivative and evaluate at x = 4.
    Last edited: Oct 14, 2006
  5. Oct 14, 2006 #4
    Ahhh alrighty, well I understand what you are saying, but I did the following:
    -- = (x+1)d/dx (squarerootx) - (squarerootx)d/dx(x+1)
    dx -------------------------------

    = (x+1)1 - (squarerootX)1

    Is that stating the same thing as you had, or have I misinterpreted this.

    Also, after that step, I'm not sure how to simplify it though. I know the (x+1)^2 moves up and the denominator becomes 2squarerootX(X+1)^2 but it's solving in the numerator that is giving me troubles.
    Last edited: Oct 14, 2006
  6. Oct 14, 2006 #5
    Actually nevermind what I just said, how abouts do you simplify the expression you just gave me courtrigad? Could you just take out the x^-0.5?
  7. Oct 14, 2006 #6
    [tex] \frac{dy}{dx} = \frac{(x+1)\frac{1}{2}x^{-\frac{1}{2}} - x^{\frac{1}{2}}}{(x+1)^{2}} = \frac{\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{1}}{x+1} = \frac{1-2x}{2\sqrt{x}(x+1)} [/tex]

    Divide by [tex] x+1 [/tex] in the numerator and denominator and change [tex] x^{-\frac{1}{2}} [/tex] to [tex] \frac{1}{\sqrt{x}} [/tex]
    Last edited: Oct 14, 2006
  8. Oct 14, 2006 #7
    Oh my goodness, how did I miss that? Haha I feel very silly now. Thank you so much, I actually just talked it over with a friend and he said the same thing. Thanks for your time though, I really appreciate it! :D
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