Question
Differentiate the following functions with respect to x:
$\frac{\text{x}}{1+\tan\text{x}}$
$\frac{\text{x}}{1+\tan\text{x}}$
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1+\tan\text{x}}\Big)$
Using quotient rule, we get
$\frac{(1+\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x})-(\text{x})\frac{\text{d}}{\text{dx}}(1+\tan\text{x})}{(1+\tan\text{x})^2}$
$=\frac{(1+\tan\text{x})-\text{x}\sec^2\text{x}}{(1+\tan\text{x})^2}$
$=\frac{1+\tan\text{x}-\text{x}\sec^2\text{x}}{(1+\tan\text{x})^2}$
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Find the middle term in the expansion of:
$\Big(\frac{2}{3}\text{x}-\frac{3}{2\text{x}}\Big)^{20}$
| Age | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 | 51-55 |
| Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
[Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]
$\frac{2\text{x}^2+\text{3x}+4}{\text{x}}$