Question

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

Answer 1

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.