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Math Help - Equation of the tangent line

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    Equation of the tangent line

    Find an equation of the tangent line to the curve at the given point. Leave your answer solved for y.

    Y=(1+x)cosx (0,1)

    Okay, so I think you just take the derivative of the function, but how do I do that and make the rest of the problem work?
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    The derivative is y'=\cos(x)-(1+x)\sin(x).
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    Quote Originally Posted by bobsanchez View Post
    Find an equation of the tangent line to the curve at the given point. Leave your answer solved for y.

    Y=(1+x)cosx (0,1)

    Okay, so I think you just take the derivative of the function, but how do I do that and make the rest of the problem work?
    Why did you suspect to take the derivative ?

    The derivative gives the slope of the tangent.
    The tangent is a straight line.
    The point (0,1) is x=0, y=1 corresponding to a point on the tangent, the point of tangency to the curve.

    To make the rest of the problem work......
    How do you write the equation of a straight line?
    If you have 2 fixed points, you can do it.
    If you have a fixed point and the slope you can do it.

    What do you do here?
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    Is the answer y = x + 1?
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    Quote Originally Posted by bobsanchez View Post
    Is the answer y = x + 1?
    Yes.

    The derivative at (0,1) is 1, so the tangent slope is 1.

    The equation of the tangent is y-1=m(x-0), so y-1=x... y=x+1
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