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Math Help - Derivitives Help!

  1. #1
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    Derivitives Help!

    Find an equation of the straight line that is tangent to the graph of f(x) root(x+1) and parallel to x - 6y+4 = 0.

    Find the slope of the tangent to the curve 1/x + 1/y = 1 at the point (2,2).
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  2. #2
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    Quote Originally Posted by mathamatics112 View Post
    Find an equation of the straight line that is tangent to the graph of f(x) = root(x+1) and parallel to x - 6y+4 = 0.
     <br />
f(x) = \sqrt{x+1}<br />

    slope of tangent = f'(x)

     <br />
f'(x) = \frac{-1}{2\sqrt{x+1}}<br />

    slope of tangent, which is parallel to line x - 6y+4 = 0 , is \frac{1}{6}

    so,

     <br />
f'(x) = \frac{-1}{2\sqrt{x+1}}=\frac{1}{6}<br />

     <br />
\sqrt{x+1}=-3<br />

     <br />
\Rightarrow x= 8<br />

     <br />
y = \sqrt{8+1}=3<br />

    so, (8, 3) is the point on the curve where tangent is drawn.

    eqn of tangent

    y = mx + b

     <br />
y = \frac{1}{6}x + b<br />

    3 = \frac{1}{6}(8) + b

     <br />
b = \frac{5}{3}<br />

    so, eqn of tangent is

     <br />
y = \frac{1}{6}x + \frac{5}{3}<br />
    Last edited by Shyam; February 15th 2009 at 06:29 PM.
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  3. #3
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    Quote Originally Posted by mathamatics112 View Post
    Find the slope of the tangent to the curve 1/x + 1/y = 1 at the point (2,2).
    \frac{1}{x}+\frac{1}{y}=1

    differentiating wrt x,

    \frac{-1}{x^2}-\frac{1}{y^2}y'=0

    y' =\frac{-y^2}{x^2}

    y' \;at\; (2, 2)  =\frac{-(2)^2}{(2)^2}=-1

    so, slope of tangent at (2, 2) is -1

    eqn of tangent

    y = mx + b

    2 = -1(2) + b

    b = 4

    eqn of tangent is

    y = -x + 4
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