Question

If the function $g\left(x\right)$ is defined by $g\left(x\right)=\frac{x^{200}}{200}+\frac{x^{199}}{199}+\frac{x^{198}}{198}+....+\frac{x^2}{2}+x+5$, then $g^{\prime }\left(0\right)=$ _____

• A. $1$
• B. $200$
• C. $100$
• D. $5$

Answer 1

Answer: (A)

$1$

We have, $g\left(x\right)=\frac{x^{200}}{200}+\frac{x^{199}}{199}+\frac{x^{198}}{198}+....+\frac{x^2}{2}+x+5$

using $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

$g^{\prime }\left(x\right)=x^{199}+x^{198}+x^{197}+.................+x+1+0$

So to get $g^{\prime }\left(0\right)$, substitute $x=0$ in above expression

$\Rightarrow g^{\prime }\left(0\right)=0+0+0+................+0+1+0=1$