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Math Help - find slope of tangent line at x=a

  1. #1
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    find slope of tangent line at x=a

    for y=1/(sqrt(x)) find slope of tangent line at the point where x=a

    then, find the equation of the tangent line at (4,1/2)
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  2. #2
    Super Member 11rdc11's Avatar
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    Take the first derivative and replace x with a to find your slope at point a.

    For the second problem do this

    f'(4)= ?

    when you find out what the slope is

    plug it into

    (y_2-y_1) = m(x_2-x_1)

    m being the slope
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mono1357 View Post
    for y=1/(sqrt(x)) find slope of tangent line at the point where x=a

    then, find the equation of the tangent line at (4,1/2)
    The slope of the tangent line at x=a is just \left.\frac{\,dy}{\,dx}\right|_{x=a}

    Thus, y=x^{-\frac{1}{2}}\implies \frac{\,dy}{\,dx}=-\tfrac{1}{2}x^{-\frac{3}{2}}.

    Thus, at x=a, \left.\frac{\,dy}{\,dx}\right|_{x=a}=-\tfrac{1}{2}a^{-\frac{3}{2}}

    Now, let's find the equation of the tangent line:

    y-f(a)=f'(a)(x-a)

    Since a=4,~f(a)=\tfrac{1}{2},~\text{and }f'(a)=-\tfrac{1}{16}, we now see that y=-\tfrac{1}{16}(x-4)+\tfrac{1}{2}\implies {\color{red}y=-\tfrac{1}{16}x+\tfrac{3}{4}}

    Does this make sense?

    --Chris
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