Proof: The Product and Quotient Rules

The product rule states that: \[\frac{d}{dx}(uv)=u \frac{dv}{dx}+v \frac{du}{dx}\] 

and the quotient rule states that: \[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v \frac{du}{dx}-u \frac{dv}{dx}}{v^2}\] where u and v are functions of x.


Proof of the Product Rule

Lets rewrite the product rule in Lagrange's notation as: \[\frac{d}{dx}(f(x)g(x))=f(x) g'(x)+g(x)f'(x)\]

Using the formula for differentiation from first principles:

\[\frac{d}{dx}(f(x)g(x))=\lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}\]


Now notice that if we add \(f(x+h)g(x)-f(x+h)g(x)\) we are adding 0 (so not changing anything), but it will allow us to factorise the fraction like so:

\[\frac{d}{dx}(f(x)g(x))=\lim_{h \to 0} \frac{f(x+h)g(x+h)+f(x+h)g(x)-f(x+h)g(x)-f(x)g(x)}{h}\]

\[\frac{d}{dx}(f(x)g(x))=\lim_{h \to 0} \frac{f(x+h)[g(x+h)+g(x)]-g(x)[f(x+h)-f(x)]}{h}\]

\[\frac{d}{dx}(f(x)g(x))=\lim_{h \to 0} \left( f(x+h) \hspace{1mm} \frac{g(x+h)+g(x)}{h}-g(x) \hspace{1mm} \frac{f(x+h)-f(x)}{h} \right)\]

\[\frac{d}{dx}(f(x)g(x))= \lim_{h \to 0} \left[  f(x+h) \hspace{1mm} \frac{g(x+h)+g(x)}{h} \right] - \lim_{h \to 0} \left[ g(x) \hspace{1mm} \frac{f(x+h)-f(x)}{h} \right]\]


\(\lim \limits_{h \to 0} f(x+h)=f(x)\) , and we can take f(x) and g(x) out of the limits like so:

\[\frac{d}{dx}(f(x)g(x))=f(x) \left[ \lim_{h \to 0} \frac{g(x+h)+g(x)}{h} \right] - g(x) \lim_{h \to 0} \left[ \frac{f(x+h)-f(x)}{h} \right]\]


Now notice that the limits are now just the formulae for differentiating f(x) and g(x) from first principles. Therefore:

\[\frac{d}{dx}(f(x)g(x))=f(x) g'(x) - g(x) f'(x)\]



Proof of the Quotient Rule

Using the product rule:

\[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{d}{dx}\left(f(x) \times \frac{1}{g(x)}\right)=f(x) \times \frac{d}{dx} \left( \frac{1}{g(x)}\right)+\frac{1}{g(x)} \times \frac{d}{dx}(f(x))\]


By the chain rule, we can see that: \[\frac{d}{dx} \left( \frac{1}{g(x)}\right)=\frac{d}{dx} ( (g(x))^{-1} )=g'(x) (-(g(x))^{-2})=-g'(x)(g(x))^{-2}=-\frac{g'(x)}{(g(x))^2}\]


Substituting this back in gives:

\[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=f(x) \times \left(-\frac{g'(x)}{(g(x))^2}\right)+\frac{1}{g(x)} \times \frac{d}{dx}(f(x))\]

\[=f(x) \times \left(-\frac{g'(x)}{(g(x))^2}\right)+\frac{1}{g(x)} \times f'(x)\]

\[=f(x) \times \left(-\frac{g'(x)}{(g(x))^2}\right)+\frac{g(x)}{(g(x))^2} \times f'(x)\]

\[=\frac{1}{(g(x))^2}\left( f(x) \times (-g'(x))+g(x) \times f'(x) \right)\]

\[=\frac{1}{(g(x))^2}\left( g(x)f'(x) -f(x)g'(x) \right)\]

\[=\frac{g(x)f'(x) -f(x)g'(x)}{(g(x))^2}\]


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