Thread: Derivative of y^2 = ax

1. 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

2. Re: Derivative of y^2 = ax

implicit is the way to go ...

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

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

3. Re: Derivative of y^2 = ax

Thank you skeeter

Originally Posted by skeeter
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.