1. ## Differentiating Implicitly!

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

2. For example, try to differentiate $3y^2$

$\frac{d}{dx}3y^2$

$\frac{d}{dx}\cdot\frac{dy}{dy}3y^2$

$\frac{dy}{dx}6y$

$6yy'$

We can generalize this to,
$\frac{d}{dx}f(y) = \frac{d}{dx}\frac{dy}{dy}f(y) = y'\cdot f'(x)$

So,

$bx^2 - p y^2 = m^2$

$2b x - 2p y y' = 0$

Take $y'$ from here..

3. Originally Posted by mathlete
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
You can differentiate constants - they are just zero!

$bx^2 -py^2 = m^2$

Derive y with respect to x using implicit differentiation. ..

$bx^2 -py^2 = m^2 \Rightarrow 2bx - 2py \frac{dy}{dx} = 0$
$py\frac{dy}{dx} = bx$
$\frac{dy}{dx} = \frac{bx}{py}$

Substitute y back into the equation and you are golden!

$bx^2 -py^2 = m^2$
$py^2 = bx^2 - m^2$
$y^2 = \frac{bx^2 - m^2}{p}$
$y = \pm \sqrt{\frac{bx^2 - m^2}{p}}$

$\frac{dy}{dx} = \frac{bx}{\pm p\sqrt{\frac{bx^2 - m^2}{p}}}$

Let's check out work another way...

$bx^2 -py^2 = m^2$

Solve for y...

$y = \pm \left(\frac{bx^2 - m^2}{p}\right)^{0.5}$

Take the derivative with respect to x

$y' = \pm \frac{2bx}{2p}\left(\frac{bx^2 - m^2}{p}\right)^{-0.5} \Rightarrow \pm \frac{bx}{p \sqrt{\frac{bx^2 - m^2}{p}}}$