Question
Find the derivative of $\text{x}^\text{n}+\text{ax}^\text{n-1}+\text{a}^2\text{x}^\text{n-2}+...+\text{a}^\text{n-1}\text{x}+\text{a}^\text{n}$ for some fixed real number a.

Answer

Let $\text{f}(\text{x})=\text{x}^\text{n}+\text{ax}^\text{n-1}+\text{a}^2\text{x}^\text{n-2}+...+\text{a}^\text{n-1}\text{x}+\text{a}^\text{n}$ $\therefore\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\big(\text{x}^\text{n}+\text{ax}^\text{n-1}+\text{a}^2\text{x}^\text{n-2}+...+\text{a}^\text{n-1}\text{x}+\text{a}^\text{n}\big)$ $=\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})+\text{a}\frac{\text{d}}{\text{dx}}(\text{x}^\text{n-1})+\text{a}^2\frac{\text{d}}{\text{dx}}(\text{x}^\text{n-2})+...+\text{a}^\text{n-1}\frac{\text{d}}{\text{dx}}(\text{x})+\text{a}^\text{n}\frac{\text{d}}{\text{dx}}(1)$ On using theorem $\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}=\text{nx}^{\text{n}-1}$, we obtain $\text{f}'(\text{x})=\text{n}\text{x}^\text{n-1}+\text{a}(\text{n-1})\text{x}^\text{n-2}+\text{a}^2(\text{n}-2)\text{x}^\text{n-3}+...+\text{a}^\text{n-1}+\text{a}^\text{n}(0)$ $=\text{n}\text{x}^\text{n-1}+\text{a}(\text{n-1})\text{x}^\text{n-2}+\text{a}^2(\text{n}-2)\text{x}^\text{n-3}+...+\text{a}^\text{n-1}$

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