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

The key idea is the derivative of a polynomial term-by-term, using the power rule $\frac{d}{dx} x^k = k x^{k-1}$.

The given expression is a finite geometric series in powers of $x$, with $n+1$ terms:

Differentiate each term separately:

• Derivative of $x^n$ is $n x^{n-1}$.
• Derivative of $a x^{n-1}$ is $a (n-1) x^{n-2}$.
• Continuing, the derivative of $a^{k} x^{n-k}$ is $a^k (n-k) x^{n-k-1}$, for $k = 0, 1, \dots, n-1$. …

Concept and Intuition

The expression $x^n + ax^{n-1} + a^2 x^{n-2} + \cdots + a^{n-1}x + a^n$ is a finite geometric series with $n+1$ terms. The first term is $x^n$, and each subsequent term multiplies the previous by $\frac{a}{x}$ (check: $x^n \cdot \frac{a}{x} = a x^{n-1}$, then $a x^{n-1} \cdot \frac{a}{x} = a^2 x^{n-2}$, and so on). The last term is $a^n$.

You could differentiate term-by-term directly — that's perfectly valid and gives the answer immediately. But the problem likely expects you to recognise the series structure, sum it, and then differentiate. Why? Because summing first often reveals a simpler function to differentiate, and it's a powerful technique for handling long or infinite series.

Let's do both: first the direct term-by-term approach (which is straightforward here), then the summation approach (which confirms the result and builds deeper understanding).

Step-by-Step Solution

1. Recognise the pattern

The given polynomial has $n+1$ terms. Write it as:

Each term is of the form $a^k x^{n-k}$ for $k = 0, 1, 2, \dots, n$. So:

2. Differentiate term-by-term (direct method)

The derivative of $a^k x^{n-k}$ with respect to $x$ is $a^k \cdot (n-k) x^{n-k-1}$, provided $n-k \geq 1$ (i.e., $k \leq n-1$). The last term ($k=n$) is $a^n$, a constant, whose derivative is $0$.

So:

Writing this out explicitly:

The last term is $a^{n-1}$ (since $x^0 = 1$). This is the derivative.

3. Alternative: Sum the series first, then differentiate

This is a geometric series with first term $x^n$ and common ratio $r = \frac{a}{x}$. The sum of $n+1$ terms is:

This holds for $x \neq a$, but since we're differentiating, we consider it as a rational function identity valid for all $x$ (the polynomial equality holds everywhere, including at $x=a$, by continuity).

Now differentiate $S(x) = \frac{x^{n+1} - a^{n+1}}{x - a}$ using the quotient rule:

Simplify the numerator: …