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Math Help - Differential and finding an equation.

  1. #1
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    Differential and finding an equation.

    Find an equation of the tangent line to the curve at the given point.
     y = (1+2x)^{10} , (0,1)

    So I use the chain rule for this don't I?

    y' = 10(1+2x)^{9}(2)

    Pretty much stuck from here.
    Thanks in advance.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cloudzer0 View Post
    Find an equation of the tangent line to the curve at the given point.
     y = (1+2x)^{10} , (0,1)

    So I use the chain rule for this don't I?

    y' = 10(1+2x)^{9}(2)

    Pretty much stuck from here.
    Thanks in advance.
    From here, substitute x=0 into the derivative:

    y^{/}(0)=20

    Plug the slope and the point into the point-slope form of a line:

    y-(1)=(20)(x-(0))=20x+1.

    Hope this helped you out!
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by cloudzer0 View Post
    Find an equation of the tangent line to the curve at the given point.
     y = (1+2x)^{10} , (0,1)

    So I use the chain rule for this don't I?

    y' = 10(1+2x)^{9}(2)

    Pretty much stuck from here.
    Thanks in advance.
    So now we need to the slope.

    To find the slope we evaluate the derivative at x=0

    m=y' = 10(1+2(0))^{9}(2)=20

    Now we use the point slope formula

    y-1=20(x-0) \iff y=20x+1

    Yeah!
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