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