Question
Differentiate the following function with respect to x:
$\text{x}^{-3}(5+\text{3x})$
$\text{x}^{-3}(5+\text{3x})$
Then, $\text{u}'=-\text{3x}^{-4};\text{v}'=3$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[\text{x}^{-3}(5+\text{3x})]=\text{x}^{-3}.3+(5+\text{3x})(-3\text{x}^{-4})$
$=\text{3x}^{-3}-\text{15x}^{-4}-\text{9x}^{-3}$
$=-\text{15x}^{-4}-\text{6x}^{-3}$
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Show that the expansion $\Big(\text{x}^2+\frac{1}{\text{x}}\Big)^{12}$ does not contain any term involving x-1.
| xi | 3 | 8 | 13 | 18 | 23 |
| fi | 7 | 10 | 15 | 10 | 6 |