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Math Help - Need help to understand how the slope was derived in this problem

  1. #1
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    Need help to understand how the slope was derived in this problem

    I am kind of struggling to understand how the slope is derived in the following problem:
    "Find the equation of the tangent line to the curve y = x^2 through the point (7, 49)."

    The answer is that the slope is 2x but I don't understand how they got there since there is only one set of points given.

    Thank you for the help,

    Alex
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  2. #2
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    y'=\mbox{exponent}*x^{\mbox{exponent -1}}

    y'=2x^{2-1}
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  3. #3
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    The slope at any point \displaystyle x on the curve \displaystyle y = x^2 is given by \displaystyle \frac{dy}{dx} = 2x.

    So at the point \displaystyle x = 7, the slope is \displaystyle 2\cdot 7 = 14.


    Equation of tangent: \displaystyle y = mx + c.

    You have a coordinate that lies on your tangent \displaystyle (x, y) = (7, 49) and the slope \displaystyle m = 14. Substitute and solve for \displaystyle c, then you can write the equation of your tangent line.
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  4. #4
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    How well do you understand the concept behind differentiation?

    If y=x^2,

    \displaystyle\frac{dy}{dx} = 2x

    <br />
\displaystyle\frac{dy}{dx} is 'the rate of change of y with respect to x'. It is used to tell us the equation of the gradient of the curve at any point.

    In the question, \frac{dy}{dx} = 2x. So take any point on your curve. Let's say (3,9) This has x co-ordinate 3.

    Substituting this into \frac{dy}{dx} = 2x gives \frac{dy}{dx} = 6. What this means is that at the point (3,9) the gradient of the curve y=x^2 is 6.

    Also, I don't think that 2x is the answer to that question.

    We have a point (7,49) and we know that \frac{dy}{dx} = 2x. So the gradient of the tangent is 2\times 7 = 14.

    So we know that when x = 7, y = 14 and m is 14. Substitute this into y = mx + c to find c and you'll find the equation of the tangent.
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  5. #5
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    from first principles ...

    \displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

    f(x) = x^2

    f(x+h) = (x+h)^2

    \displaystyle f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}

    \displaystyle f'(x) = \lim_{h \to 0} \frac{x^2+2xh+h^2 - x^2}{h}

    \displaystyle f'(x) = \lim_{h \to 0} \frac{2xh+h^2}{h}<br />

    \displaystyle f'(x) = \lim_{h \to 0} \frac{h(2x+h)}{h}<br />

    \displaystyle f'(x) = \lim_{h \to 0} 2x+h = 2x
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