Question
Differentiate the following from the first principle
$\sqrt{\sin\text{2x}}$
$\sqrt{\sin\text{2x}}$
$\text{f}(\text{x})=\sqrt{\sin\text{2x}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sqrt{\sin\text{2x}}}{\text{h}}$
Multiplying numerator and denominator by
$\Big(\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}\Big)$$\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sqrt{\sin\text{2x}}}{\text{h}}\times\frac{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}}{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin2\text{x}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h})-\sin\text{2x}}{\Big(\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin2\text{x}}\Big)}$ $\Big[\sin\text{c}-\sin\text{d}=2\cos\frac{\text{c}+\text{d}}{2}\sin\frac{\text{c}-\text{d}}{2}\Big]$
$\lim_\limits{\text{h}\rightarrow0}\frac{2\cos(\text{2x}+\text{h})\times\sin\text{h}}{\text{h}}\times\frac{1}{\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin\text{2x}}}$
$=\frac{2\cos\text{2x}}{2\sqrt{\sin\text{2x}}}$
$=\frac{\cos\text{2x}}{\sqrt{\sin\text{2x}}}$
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| Subject | Mathematics | Physics | Chemistry |
| Mean | 42 | 32 | 40.9 |
| Standard deviation | 12 | 15 | 20 |
Which of these three subjects shows the highest variability in marks and which shows the lowest?
| Column C1 | Column C2 | ||
| (a) | $\sin(\text{x + y})\sin\text{x}-\text{y}$ | (i) | $\cos^2\text{x}-\sin^2\text{y}$ |
| (b) | $\cos(\text{x + y})\cos(\text{x}-\text{y})$ | (ii) | $\frac{1-\tan\theta}{1+\tan\theta}$ |
| (c) | $\cot\Big(\frac{\pi}{4}+\theta\Big)$ | (iii) | $\frac{1+\tan\theta}{1-\tan\theta}$ |
| (d) | $\tan\Big(\frac{\pi}{4}+\theta\Big)$ | (iv) | $\sin^2\text{x}-\sin^2\text{y}$ |