d/dx (u/v)

burgerandcheese

Junior Member
Joined
Jul 2, 2018
Messages
73
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

12116
 

Romsek

Full Member
Joined
Nov 16, 2013
Messages
358
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
Joined
Jul 2, 2018
Messages
73
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
 
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