WEBVTT
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suppose you want to find an equation of the tangent
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line to the curve. Why? Which is equal
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to two? X plus one over X plus two
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. At the.1, 1 to do this,
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we first find the slope of the tangent line.
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That is the derivative of the function Evaluated at the
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given.1, 1. Now by quotient rule we
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have white prime, that's equal to X plus two
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times the derivative of the numerator two, x plus
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one-. We have to express one times the
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derivative of the denominator which is expressed to this all
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over the square of the denominator. And then from
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here we have X plus two Times derivative of to
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Express one. That's just too minus two, X
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plus one times the derivative of X plus two which
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is one. And then this all over the square
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of X plus two. And simplifying this, we
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have two, X plus four minus two, X
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-1. This all over X plus two squared or
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this is just three over the square of experts to
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And so the slope of the tangent line at the
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.11 is given by that's dy over dx evaluated at
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1 1. This is just three over one plus
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two squared. That's just 3/9 or 1/3. So
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this is the slope of the tangent line at 11
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. And then the next step would be to use
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the point slope formula of a line to find the
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equation of the tangent line. Now the point slope
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formula of a line states that the the equation of
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the line is just why minus? Why is that
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one? This is equal to the slope mm Of
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that line times X-X. Sub one. Since
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you already have our point Except one White 1 Which
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is just 11. And we have our slope,
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we found out To be won over three. Then
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the equation of the tangent line is just Why-1
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That's equal to 1/3 times x-1. And simplifying
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this, we have why that's equal to 1/3,
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X minus 1/3 Plus one. Or that why is
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equal to 1/3 Plus 2/3. And so this is
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the equation of the tangent line at the point 11