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📐 Mathematics  ·  Limits and Derivatives  ·  JEE

Taylor series expansion of eˣ around x=0 gives the n-th derivative value at 0 as:

Answer: 1.

  • A n!
  • B 1
  • C n
  • D 0

Correct answer: B. 1

Explanation: Taylor series: eˣ = Σ(xⁿ/n!). The coefficient of xⁿ is f⁽ⁿ⁾(0)/n!, and here it equals 1/n!. So f⁽ⁿ⁾(0) = 1 for all n ≥ 0, confirmed since d<sup>n</sup>/dx<sup>n</sup>(eˣ) = eˣ and e⁰ = 1.

xyP (a, f(a))Q (b, f(b))secant PQtangent at PAs Q slides toward P, secant approaches tangent

As point Q slides along the curve toward P, the secant line PQ rotates into the tangent line at P, whose slope is the derivative.

Concept context

Rate of change, differentiation rules, and applications

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