Question

If y = 1 +$\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+......,\mathrm{then}\frac{dy}{dx}=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

If y = 1 + $\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+......$
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+......\right) \\ =\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(\frac{x}{1!}\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right)+\frac{d}{dx}\left(\frac{x^3}{3!}\right)+.... \\ =0+\frac{1}{1!}\frac{d}{dx}\left(x\right)+\frac{1}{2!}\frac{d}{dx}\left(x^2\right)+\frac{1}{3!}\frac{d}{dx}\left(x^3\right)+..... \\ =0+\frac{1}{1!}\times 1+\frac{1}{2!}\times 2x+\frac{1}{3!}\times 3x^2+...... \\ =1+\frac{x}{1!}+\frac{1}{3\times 2!}3x^2+\frac{1}{4\times 3!}4x^3+..... \\ \mathrm{i}.\mathrm{e}\frac{dy}{dx}=1+\frac{x}{1!}+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+...... \end{gathered}$$