Question

How do you find the slope of the tangent line `y=(x+1)(x-2)` at x=1?

Answer 1

The slope of the tangent line to `y` is `1`

Explanation:

The slope of the tangent line is evaluated by computing `y'_(x=1)`
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Let us differentiate `y`:
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Applying the product rule of differentiation :

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`color(blue)((uv)'=u'v + v'u`
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`color(blue)(y'=(x+1)'(x-2)+(x-2)'(x+1)`
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`y'=1(x-2)+1(x+1)`
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`y'=x-2+x+1`
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`y'=2x-1`
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The slope is `y'_(x=1)`, so let us substitute `x=1` in `y'`

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`y'_(x=1) =2(1)-1`

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`y'_(x=1) =2-1=1`
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Therefore, the slope of the tangent line at `x=1" "` is `1`.