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  1. #1 Calc Question 
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    Can any of you guys walk me through what to do on #16? I know how to find the derivative, but am clueless as to how I would relate it to finding a tangent line.

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    Please keep in mind that the derivative is the equation for the slope of the tangent line. To form the equation of the line you need a point and a slope.

    It's asking you to find the equation of the tangent line given an equation and a point.
    Start off by finding out what the point is by plugging in x into the equation.
    x = 0
    y = e^(0+1) y = 2.72

    pt = (0, 2.72)
    y' = e^(x+1)


    Remembering that the derivative is the slope of the tangent line allows us to plug x into the derivative to get the slope of the tangent line at that point.
    f'(0) = 2.72
    m = 2.72


    Now, using the point slope formula you can create an equation for the tangent line:
    y-y1= m(x-x1)
    where y1 and x1 are the point you found and m is the slope you calculated.

    y-2.72 = 2.72(x-0)
    y-2.72 = 2.72x
    y = 2.72x + 2.72


    Please correct me if I went wrong anywhere.


    EDIT: Please note 2.72 = e^1
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    Quote Originally Posted by All3n View Post
    Please keep in mind that the derivative is the equation for the slope of the tangent line. To form the equation of the line you need a point and a slope.

    It's asking you to find the equation of the tangent line given an equation and a point.
    Start off by finding out what the point is by plugging in x into the equation.
    x = 0
    y = e^(0+1) y = 2.72

    pt = (0, 2.72)
    y' = e^(x+1)


    Remembering that the derivative is the slope of the tangent line allows us to plug x into the derivative to get the slope of the tangent line at that point.
    f'(0) = 2.72
    m = 2.72


    Now, using the point slope formula you can create an equation for the tangent line:
    y-y1= m(x-x1)
    where y1 and x1 are the point you found and m is the slope you calculated.

    y-2.72 = 2.72(x-0)
    y-2.72 = 2.72x
    y = 2.72x + 2.72


    Please correct me if I went wrong anywhere.
    Thanks so much man. For some reason I forgot the derivative was the slope of the tangent line haha. So much clearer now, cheers
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  5. #4  
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    An easy way to remember is to just think of the derivative as a way of saying the "rate of change" of a function. If you made a bunch of tangent lines on your original function and proceeded to calculate their slopes, the same effect can be modelled by taking the derivative and producing a general equation to represent this procedure/act. Remembering the non calculus equivalent and the true definition of a derivative (have you done this yet?) helps make things intuitive.
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  6. #5  
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    Quote Originally Posted by Numbers View Post
    An easy way to remember is to just think of the derivative as a way of saying the "rate of change" of a function. If you made a bunch of tangent lines on your original function and proceeded to calculate their slopes, the same effect can be modelled by taking the derivative and producing a general equation to represent this procedure/act. Remembering the non calculus equivalent and the true definition of a derivative (have you done this yet?) helps make things intuitive.
    Thanks for the help man. Sometimes I just get tripped up in the problem and overlook the logicality of the problem.
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