proof of quotient rule

Let $F(x)=f(x)/g(x)$. Then

$F^{\prime }(x)$ $=$ $\underset{h\to 0}{lim}{\displaystyle \frac{F(x+h)-F(x)}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}}$

$=$ $\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x)-f(x)g(x+h)}{hg(x+h)g(x)}}$

Like the product rule, the key to this proof is subtracting and adding the same quantity. We separate $f$ and $g$ in the above expression by subtracting and adding the term $f(x)g(x)$ in the numerator.

$F^{\prime }(x)$ $=$ $\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{hg(x+h)g(x)}}$

$=$ $\underset{h\to 0}{lim}{\displaystyle \frac{g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x)}{h}}{g(x+h)g(x)}}$

$=$ ${\displaystyle \frac{{lim}_{h\to 0}g(x)\cdot {lim}_{h\to 0}\frac{f(x+h)-f(x)}{h}-{lim}_{h\to 0}f(x)\cdot {lim}_{h\to 0}\frac{g(x+h)-g(x)}{h}}{{lim}_{h\to 0}g(x+h)\cdot {lim}_{h\to 0}g(x)}}$

$=$ ${\displaystyle \frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{{[g(x)]}^2}}$

Title	proof of quotient rule
Canonical name	ProofOfQuotientRule
Date of creation	2013-03-22 12:38:58
Last modified on	2013-03-22 12:38:58
Owner	drini (3)
Last modified by	drini (3)
Numerical id	5
Author	drini (3)
Entry type	Proof
Classification	msc 26A06