# proof of quotient rule

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

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

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.

 $\displaystyle F^{\prime}(x)$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{hg(x+% h)g(x)}$ $\displaystyle=$ $\displaystyle\lim_{h\to 0}\frac{g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x% )}{h}}{g(x+h)g(x)}$ $\displaystyle=$ $\displaystyle\frac{\lim_{h\to 0}g(x)\cdot\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}-% \lim_{h\to 0}f(x)\cdot\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}}{\lim_{h\to 0}g(x+h)% \cdot\lim_{h\to 0}g(x)}$ $\displaystyle=$ $\displaystyle\frac{g(x)f^{\prime}(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
Title proof of quotient rule ProofOfQuotientRule 2013-03-22 12:38:58 2013-03-22 12:38:58 drini (3) drini (3) 5 drini (3) Proof msc 26A06