Question
Differentiate the following from first principle:
$\text{x}\text{e}^\text{x}$
$\text{x}\text{e}^\text{x}$
$\text{f(x)}=\text{xe}^\text{x}$
$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x+h})\text{e}^{(\text{x+h})}-\text{xe}^\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{xe}^\text{x}.\text{e}^\text{h}+\text{he}^\text{x}.\text{e}^\text{h}-\text{xe}^\text{x}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{xe}^\text{x}\Big(\frac{\text{e}^\text{h}-1}{\text{h}}\Big)+\frac{\text{he}^{\text{x}+\text{h}}}{\text{h}}$
$=\text{xe}^\text{x}+\text{e}^\text{x}$
$=\text{e}^\text{x}\big(\text{x}+1\big)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Weight (in grams): | 200-201 | 201-202 | 202-203 | 203-204 | 204-205 | 205-206 |
| Frequency: | 13 | 27 | 18 | 10 | 1 | 1 |
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})...+(\text{a}+(\text{n}-1)\text{d})=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
| Column A | Column B | ||
| a. | The polar form of $\text{i}+\sqrt{3}$ is | i. | Perpendicular bisector of segment joining (– 2, 0) and (2, 0). |
| b. | The amplitude of $-1+\sqrt{-3}$ is | ii. | On or outside the circle having centre at (0, – 4) and radius 3. |
| c. | If |z + 2| = |z - 2|, then locus of z is | iii. | $\frac{2\pi}{3}$ |
| d. | If |z + 2i| = |z - 2i|, then locus of z is | iv. | Perpendicular bisector of segment joining (0, – 2) and (0, 2). |
| e. | Region represented by $|\text{z}+4\text{i}|\geq3$ is | v. | $2\Big(\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$ |
| f. | Region represented by $|\text{z}+4|\leq3$ is | vi. | On or inside the circle having centre (– 4, 0) and radius 3 units. |
| g. | Conjugate of $\frac{1+2\text{i}}{1-\text{i}}$ lies in | vii. | First quadrant. |
| h. | Reciprocal of 1 - i lies in | viii. | Third quadrant. |