Question

Given that $f\left(x\right)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$ up to infinite terms.

The derivative of $f\left(x\right)$ is .

• A. $f\left(x\right)$
• B. $2f\left(x\right)$
• C. $\left[f\right(x){]}^2$
• D. $\frac{f\left(x\right)}{x!}$

Answer 1

The correct option is A $f\left(x\right)$

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{1}{h}[(1+(x+h)+\frac{(x+h{)}^2}{2!}+...)-(1+x+\frac{x^2}{2!}+...)]$
$=\underset{h\to 0}{lim}\frac{1}{h}[(h+\frac{2hx}{2!}+\frac{3h{x}^2}{3!}+...)+\text{terms having}{h}^2\text{and higher powers}]=1+x+\frac{x^2}{2!}+....=f\left(x\right)$