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  4. If the function g(x) is defined by g(x)= (x200/200)+ (x199/199)+ (x198 /198)+.....+ (x2/2)+x+5 then g'(0) =

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Q. If the function $g(x)$ is defined by $g(x)= \frac {x^{200}}{200}+ \frac {x^{199}}{199}+\frac {x^{198}} {198}+.....+\frac {x^2}{2}+x+5$ then $g'(0)$ =

KCETKCET 2015Limits and Derivatives

Solution:

We have,
$ g(x)=\frac{x^{200}}{200}+\frac{x^{199}}{199}+\frac{x^{198}}{198}+\ldots+\frac{x^{2}}{2}+x+5 $
$\Rightarrow g'(x)=x^{199}+x^{198}+x^{197}+\ldots+x+1$
$\Rightarrow g'(0)=0+0+0+\ldots+0+1=1$