Question
Differentiate the following from the first principle
$\sin(2\text{x}-3)$
$\sin(2\text{x}-3)$
$\text{f}(\text{x})=\sin(\text{2x}-3)$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\big\{2(\text{x}+\text{h})-3\big\}-\sin(\text{2x}-3)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\cos\frac{(\text{2x}+\text{2h}-3+\text{2x}-3)}{2}\times\sin\frac{(\text{2x}+\text{2h}-3-2-\text{x}+3)}{2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}2\cos(\text{2x}-3+\text{h}).\frac{\sin\text{x}}{\text{x}}$
$=2\cos(\text{2x}-3)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Size (in cms): | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
| No. of items: | 2 | 8 | 20 | 35 | 20 | 15 |
$\text{B}'\subset\text{A}'\Rightarrow\text{A}\subset\text{B.}$
If the term from x in the expansion of $\Big(\sqrt{\text{x}}-\frac{\text{k}}{\text{x}^{2}}\Big)^{10}$ is 405, find the value of k.
| Year render: | 10 | 20 | 30 | 40 | 50 | 60 |
| No. of persons(cumulative): | 15 | 32 | 51 | 78 | 97 | 109 |
| Number | Boys | Girls |
| 100 | 50 | |
| Mean weight | 60kg | 45kg |
| Variance | 9 | 4 |