Find the derivative of function y = 1 + x + x2 + …. + x50 at x = 1.

1.	Find the derivative of function y = 1 + x + x2 + …. + x50 at x = 1.
Answer» Given, \(y=1+x+x^2+...+x^{50}.\) ∴ \(\frac{dy}{dx}=\frac{d}{dx}[1+x+x^2+...+x^{50}]\) \(=\frac{d}{dx}1+\frac{d}{dx}x+\frac{d}{dx}x^2+...+\frac{d}{dx}x^{50}\) ⇒ \(\frac{dy}{dx}\)= 1+2x+3x²+....+50x⁴⁹ ∴ At x = 1, \(\frac{dy}{dx}\)= 1+2+3+....+50 = \(\frac{50(50+1)}{2}\) = 25 × 51 = 1275.

Answer»

Given,

\(y=1+x+x^2+...+x^{50}.\)

∴ \(\frac{dy}{dx}=\frac{d}{dx}[1+x+x^2+...+x^{50}]\)

\(=\frac{d}{dx}1+\frac{d}{dx}x+\frac{d}{dx}x^2+...+\frac{d}{dx}x^{50}\)

⇒ \(\frac{dy}{dx}\)= 1+2x+3x²+....+50x⁴⁹

∴ At x = 1,

\(\frac{dy}{dx}\)= 1+2+3+....+50

= \(\frac{50(50+1)}{2}\)

= 25 × 51

= 1275.

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