d/dx (u/v)

burgerandcheese

Junior Member
Hi

In the picture below, ignore the part I crossed out with red. How did they get the -1/v2 (underlined in blue)? Could it have been from the g'(x) above? I still can't see how the writer got it.

If I use chain rule to differentiate u/v,

d/dx (u/v) = u*dv/dx + (1/v)*du/dx

Romsek

Full Member
They are applying the product rule to $$\displaystyle \dfrac{d}{dx}\left(u \times \dfrac 1 \nu\right)$$

$$\displaystyle \dfrac{d}{dx}\left(u \times \dfrac 1 \nu\right) = \dfrac{du}{dx}\times \dfrac 1 \nu + u \times \dfrac{d}{dx}\left(\dfrac 1 \nu\right)$$

$$\displaystyle \dfrac{d}{dx}\left(\dfrac 1 \nu\right) = \dfrac{d}{d\nu}\left(\dfrac 1 \nu\right)\dfrac{d\nu}{dx} = \left(-\dfrac{1}{\nu^2}\right) \dfrac{d\nu}{dx}$$

burgerandcheese

Junior Member
They are applying the product rule to $$\displaystyle \dfrac{d}{dx}\left(u \times \dfrac 1 \nu\right)$$

$$\displaystyle \dfrac{d}{dx}\left(u \times \dfrac 1 \nu\right) = \dfrac{du}{dx}\times \dfrac 1 \nu + u \times \dfrac{d}{dx}\left(\dfrac 1 \nu\right)$$

$$\displaystyle \dfrac{d}{dx}\left(\dfrac 1 \nu\right) = \dfrac{d}{d\nu}\left(\dfrac 1 \nu\right)\dfrac{d\nu}{dx} = \left(-\dfrac{1}{\nu^2}\right) \dfrac{d\nu}{dx}$$
Wow, thank you