Find dy/dx. Assume b, p, and m are constants.
bx^2 -py^2 = m^2
i need some help with this, i know you cant differentiate constants, so how would i do this one...
mathlete
For example, try to differentiate $\displaystyle 3y^2$
$\displaystyle \frac{d}{dx}3y^2$
$\displaystyle \frac{d}{dx}\cdot\frac{dy}{dy}3y^2$
$\displaystyle \frac{dy}{dx}6y$
$\displaystyle 6yy'$
We can generalize this to,
$\displaystyle \frac{d}{dx}f(y) = \frac{d}{dx}\frac{dy}{dy}f(y) = y'\cdot f'(x)$
So,
$\displaystyle bx^2 - p y^2 = m^2$
$\displaystyle 2b x - 2p y y' = 0$
Take $\displaystyle y'$ from here..
You can differentiate constants - they are just zero!
$\displaystyle bx^2 -py^2 = m^2$
Derive y with respect to x using implicit differentiation. ..
$\displaystyle bx^2 -py^2 = m^2 \Rightarrow 2bx - 2py \frac{dy}{dx} = 0$
$\displaystyle py\frac{dy}{dx} = bx$
$\displaystyle \frac{dy}{dx} = \frac{bx}{py}$
Substitute y back into the equation and you are golden!
$\displaystyle bx^2 -py^2 = m^2$
$\displaystyle py^2 = bx^2 - m^2$
$\displaystyle y^2 = \frac{bx^2 - m^2}{p}$
$\displaystyle y = \pm \sqrt{\frac{bx^2 - m^2}{p}}$
$\displaystyle \frac{dy}{dx} = \frac{bx}{\pm p\sqrt{\frac{bx^2 - m^2}{p}}}$
Let's check out work another way...
$\displaystyle bx^2 -py^2 = m^2$
Solve for y...
$\displaystyle y = \pm \left(\frac{bx^2 - m^2}{p}\right)^{0.5}$
Take the derivative with respect to x
$\displaystyle y' = \pm \frac{2bx}{2p}\left(\frac{bx^2 - m^2}{p}\right)^{-0.5} \Rightarrow \pm \frac{bx}{p \sqrt{\frac{bx^2 - m^2}{p}}}$