I need HELP!! T_T

Do I have to use the ln fact? u(x)/v(x) is like saying u(x) times v(x)^-1. For simplicity, I'll use u for u(x) and v for v(x) Now use the product rule:

d/dx [u * v^-1]

d/dx [u] * v^-1 + u * d/dx [v^-1] Use chain rule

u' * v​^-1 + u * (-1)(v)^-2 * v' 

u'/v - uv'/v²  Multiply u'/v by v to combine into a single denominator and:

(u'v - uv')/v² 

 

So:

d/dx [u/v] = (u'v - uv')/v² 

 Using the fact that "ln ( u(x)/v(x) )= ln u(x)−ln v(x),

use the chain rule and the formula for the derivative of ln x to derive the quotient rule. That is, find the derivative of [u(x)/v(x)] without using the quotient rule.  

 

Thanks HamboneJoe,

Here is an alternate answer using the given fact.

 

\(let\;u=u(x), \quad v=v(x), \quad \frac{du}{dx}=u', \quad \frac{dv}{dx}=v'\\~ \\ ln\;\frac{u}{v}=lnu-lnv\\ e^{ln\;\frac{u}{v}}=e^{(lnu-lnv)}\\ \frac{u}{v}=e^{(lnu-lnv)}\\ \frac{d}{dx}\frac{u}{v}=\frac{d}{dx}(lnu-lnv)*e^{(lnu-lnv)}\\ \frac{d}{dx}\frac{u}{v}=(\frac{u'}{u}-\frac{v'}{v})*e^{ln(\frac{u}{v})}\\ \frac{d}{dx}\frac{u}{v}=\frac{vu'-uv'}{uv}*\frac{u}{v}\\ \frac{d}{dx}\frac{u}{v}=\frac{vu'-uv'}{v^2}\\ \)