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Thread: [SOLVED] Differentiation Question

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    [SOLVED] Differentiation Question

    Question:
    The equation of a curve is $\displaystyle y = \frac{6}{5 - 2x}$
    (i) Calculate the gradient of the curve at the point where $\displaystyle x = 1$
    (ii) A point with coordinates $\displaystyle (x, y)$ moves along the curve in such a way that the rate of increase of $\displaystyle y$ has a constant value of $\displaystyle 0.02$ units per second. Find the rate of increase of $\displaystyle x$ when $\displaystyle x = 1$.


    Answers:

    (i) $\displaystyle y = 6(5 - 2x)^{-1}$
    $\displaystyle \frac{dy}{dx} = 6 \times -1(5 - 2x)^{-1-1} \times (-2)$
    $\displaystyle \frac{dy}{dx} = 12(5 - 2x)^{-2}$

    Is this right?

    (ii) How can I get the rate of increase of $\displaystyle x$?
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    Hello ^^

    Quote Originally Posted by looi76 View Post
    Question:
    The equation of a curve is $\displaystyle y = \frac{6}{5 - 2x}$
    (i) Calculate the gradient of the curve at the point where $\displaystyle x = 1$
    (ii) A point with coordinates $\displaystyle (x, y)$ moves along the curve in such a way that the rate of increase of $\displaystyle y$ has a constant value of $\displaystyle 0.02$ units per second. Find the rate of increase of $\displaystyle x$ when $\displaystyle x = 1$.


    Answers:

    (i) $\displaystyle y = 6(5 - 2x)^{-1}$
    $\displaystyle \frac{dy}{dx} = 6 \times -1(5 - 2x)^{-1-1} \times (-2)$
    $\displaystyle \frac{dy}{dx} = 12(5 - 2x)^{-2}$

    Is this right?
    Yes it is
    Remember, you're asked the gradient where x=1

    (ii) How can I get the rate of increase of $\displaystyle x$?
    The rate of increase is 0.02 means that $\displaystyle \frac{d^2y}{dx^2}=0.02$ (second derivative)

    The second derivative represents the speed of variation of y, so if you're asked a rate of increasing or decreasing, use this
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