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Thread: Derivative of y^2 = ax

  1. #1
    Member Furyan's Avatar
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    Derivative of y^2 = ax

    Hello

    If I want to find the gradient at any point on a parabola with an equation of the form:

    $\displaystyle y^2 = ax$ , where $\displaystyle a > 0$

    Do I use $\displaystyle \dfrac{dy}{dx} = \dfrac{a}{2\sqrt{ax}}$

    knowing that when $\displaystyle y > 0, \dfrac {dy}{dx} > 0$ and that when $\displaystyle y < 0, \dfrac {dy}{dx} < 0$ or is there another way? I noticed when I used implicit differentiation I didn't get an $\displaystyle x$ term in the derivative.

    Thank you
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  2. #2
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    Re: Derivative of y^2 = ax

    implicit is the way to go ...

    $\displaystyle 2y \cdot \frac{dy}{dx} = a$

    $\displaystyle \frac{dy}{dx} = \frac{a}{2y}$
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  3. #3
    Member Furyan's Avatar
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    Re: Derivative of y^2 = ax

    Thank you skeeter

    Quote Originally Posted by skeeter View Post
    implicit is the way to go ...

    $\displaystyle 2y \cdot \frac{dy}{dx} = a$

    $\displaystyle \frac{dy}{dx} = \frac{a}{2y}$
    I replaced my $\displaystyle 2y$ with $\displaystyle 2\cdot\dfrac{dy}{dx}$ instead of $\displaystyle 2y \cdot \frac{dy}{dx}$, so I didn't have a $\displaystyle y$ term in the derivative either. That's why I was confused.

    Thank you very much and a very happy New Year to you.
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