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Worked Examples · Example 14

Q.Find the derivative of f(x)=1+x+x2+x3+⋯+x50f(x) = 1 + x + x^2 + x^3 + \cdots + x^{50} at x=1x = 1.

Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
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✓ Free question

The derivative of the sum f(x)=1+x+x2+⋯+x50f(x) = 1 + x + x^2 + \cdots + x^{50} at x=1x=1 is found by differentiating term-by-term and evaluating the resulting arithmetic series. The value is 1275\boxed{1275}.

The key insight: when you have a polynomial written as a sum of powers, the derivative is simply the sum of the derivatives of each term. At x=1x=1, each derivative term nxn−1n x^{n-1} becomes just nn, so the problem reduces to adding the numbers from 1 to 50.

  1. Differentiate term by term. The derivative of a constant is 0. For each xnx^n where n≥1n \ge 1, we have ddxxn=nxn−1\frac{d}{dx} x^n = n x^{n-1}. So:

f′(x)=0+1⋅x0+2⋅x1+3⋅x2+⋯+50⋅x49f'(x) = 0 + 1 \cdot x^{0} + 2 \cdot x^{1} + 3 \cdot x^{2} + \cdots + 50 \cdot x^{49}

That is:

f′(x)=1+2x+3x2+⋯+50x49f'(x) = 1 + 2x + 3x^2 + \cdots + 50x^{49}

  1. Evaluate at x=1x = 1. At x=1x = 1, every power of xx becomes 1. So:

f′(1)=1+2+3+⋯+50f'(1) = 1 + 2 + 3 + \cdots + 50

  1. Sum the arithmetic series. The sum of the first nn natural numbers is n(n+1)2\frac{n(n+1)}{2}. Here n=50n = 50:

f′(1)=50×512=25×51=1275f'(1) = \frac{50 \times 51}{2} = 25 \times 51 = 1275

Watch out

A common mistake is to forget that the constant term 11 differentiates to 00, not to 11. Also, students sometimes stop at the expression 1+2+⋯+501 + 2 + \cdots + 50 without actually summing it — in an exam, you must compute the numeric value.

Tip

If the series went up to x50x^{50}, the derivative at x=1x=1 is always the sum 1+2+⋯+501 + 2 + \cdots + 50. For a general upper limit NN, the answer is N(N+1)2\frac{N(N+1)}{2}. This pattern saves time in multiple-choice questions.

✓Final answer

The derivative at x=1x = 1 is 1275\boxed{1275}.

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