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NCERT Exemplar · Q79

Q.If y=1+x1!+x22!+x33!+...y = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ..., then dydx=\dfrac{dy}{dx} = ________.

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The series is the exponential function exe^x, whose derivative is itself; thus dydx=ex=y\frac{dy}{dx} = e^x = y.

The series you're looking at is one of the most beautiful in mathematics. Before we differentiate term-by-term, recognize what this function actually is.

Why this series matters

The infinite series

y=1+x1!+x22!+x33!+⋯=∑n=0∞xnn!y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}

is the Taylor series expansion of exe^x about x=0x = 0. This is not just any series—it's the defining series for the exponential function, convergent for all real xx.

The key insight: when we differentiate exe^x, we get exe^x back. Let's see why that emerges naturally from the series itself.

Differentiating term by term

Power series can be differentiated term-by-term within their radius of convergence (which here is infinite). Let's differentiate each term:

  1. The constant term: ddx(1)=0\frac{d}{dx}(1) = 0

  2. The linear term: ddx(x1!)=11!=1\frac{d}{dx}\left(\frac{x}{1!}\right) = \frac{1}{1!} = 1

  3. The quadratic term: ddx(x22!)=2x2!=x1!\frac{d}{dx}\left(\frac{x^2}{2!}\right) = \frac{2x}{2!} = \frac{x}{1!}

  4. The cubic term: ddx(x33!)=3x23!=x22!\frac{d}{dx}\left(\frac{x^3}{3!}\right) = \frac{3x^2}{3!} = \frac{x^2}{2!}

  5. The general term: ddx(xnn!)=nxn−1n!=xn−1(n−1)!\frac{d}{dx}\left(\frac{x^n}{n!}\right) = \frac{nx^{n-1}}{n!} = \frac{x^{n-1}}{(n-1)!}

Notice the pattern: each term becomes the previous term in the original series.

The beautiful result

Collecting all these derivatives: …

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