Question
Differentiate the following from first principle:
$\frac{\text{x}+2}{3\text{x}+5}$
$\frac{\text{x}+2}{3\text{x}+5}$
$\text{f(x)}=\frac{\text{x+2}}{3\text{x}+5}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{(x+h+2)}}{3\text{(x+h)+5}}-\frac{\text{x+2}}{3\text{x+5}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{3x+5}\big)\big(\text{x+h+2}\big)-\big(\text{x+2}\big)\big(\text{3x+3h+5}\big)}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{3x}^2+5\text{x}+3\text{xh}+5\text{h}+6\text{x}+10\big)-\big(\text{3x}^2+3\text{xh}+5\text{x}+6\text{x}+6\text{h}+10\big)}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)}$
$=\frac{-1}{\big(\text{3x+5}\big)^2}$
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$\frac{\text{a}}{1+\text{i}}+\frac{\text{a}}{(1+\text{i})^2}+\frac{\text{a}}{(1+\text{i})^3}+\ ...\ +\frac{\text{a}}{(1+\text{i})^\text{n}}.$
$(\text{a}_0\text{x}^\text{n}+\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+\dots+\text{a}_{\text{n}-1}+\text{a}_\text{n})$