Question
Differentiate the following from first principle:
$\text{x}^3-\text{5x}^2+\text{x}-5$
$\text{x}^3-\text{5x}^2+\text{x}-5$
$\text{f(x)}=\text{x}^3-5\text{x}^2+\text{x}-5$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg\{(\text{x+h})^3+(\text{x+h})-5(\text{x+h})^2-5\bigg\}-(\text{x}^3-5\text{x}^2+\text{x}-5)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}^3+\text{h}^3+3\text{x}^2\text{h}+3\text{h}^2\text{x}+\text{x}+\text{h}-5\text{x}^2-5\text{h}-10\text{xh}-5)-(\text{x}^3-5\text{x}^2+\text{x}-5)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(3\text{x}^2\text{h}+3\text{h}^2\text{x}+\text{h}^3+\text{h}-5\text{h}^2-10\text{xh})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}3\text{x}^2+3\text{xh}+\text{h}^2+1-5\text{h}-10\text{x}$
$=3\text{x}^2-10\text{x}+1$
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$\frac{\text{2x}+3}{\text{x}-2}$
If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?
| xi | 5 | 7 | 9 | 10 | 12 | 15 |
| fi | 8 | 6 | 2 | 2 | 2 | 6 |