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Math Help - tangent line or something.

  1. #1
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    tangent line or something.

    How do you find an equation of the target line to the at the given point : y=(1+x)cosx at the point 0,1.

    I think you use product rule and then plug it into y=mx+b but i cant get right answer
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: tangent line or something.

    Tangent: y - y(0)=y'(0)*(x-x0)

    {x0,y(x0)} = {0,1}

    y'(x) = cos(x) - (1+x)*sin(x)
    Last edited by MaxJasper; October 12th 2012 at 09:25 AM.
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  3. #3
    Behold, the power of SARDINES!
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    Re: tangent line or something.

    Quote Originally Posted by kaw95 View Post
    How do you find an equation of the target line to the at the given point : y=(1+x)cosx at the point 0,1.

    I think you use product rule and then plug it into y=mx+b but i cant get right answer
    Yes you use the product rule

    This gives

    y'(x)=-(1+x)\sin(x)+\cos(x)

    Now we can find the slope at the point (0,1)

    m=y'(0)=-(1+0)\sin(0)+\cos(0)=1

    Since we know the y intercept (the point (0,1) )

    We know b=1

    This gives

    y=1(x)+1 \implies y=x+1

    P.S where in Oregon are you from? I do miss Corvallis and Portland quite alot! Go Beavers!!
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    Re: tangent line or something.

    "the derivative is the slope of the tangent line"
    That's perhaps the single most important and commonly used phrase in Calculus I.
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