Question
Differentiate the following from first principle:
$\text{x}^3+4\text{x}^2+3\text{x}+2$
$\text{x}^3+4\text{x}^2+3\text{x}+2$
$\text{f(x)}=\text{x}^3+4\text{x}^2+3\text{x}+2$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x+h})^3+4(\text{x+h})^2+3(\text{x+h})+2-(\text{x}^3+4\text{x}^2+3\text{x}+2)}{\text{h}}$
On solving we get,$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^3+\text{h}^3+3\text{x}^2\text{h}+3\text{h}^2\text{x}+4\text{x}^2+4\text{h}^2+8\text{hx}+3\text{x}+3\text{h}+2-\text{x}^3-4\text{x}^2-3\text{x}-2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{3\text{x}^2\text{h}+3\text{xh}^2+\text{h}^3+4\text{h}^2+8\text{hx}+3\text{h}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}3\text{x}^2+3\text{xh}+\text{h}^2+4\text{h}+8\text{x}+3$
$=3\text{x}^2+8\text{x}+3$
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| Subject | Mathematics | Physics | Chemistry |
| Mean | 42 | 32 | 40.9 |
| Standard deviation | 12 | 15 | 20 |
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