If the function g(x) is defined by g(x) = fra | vedclass

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(B) Given the function: $g(x) = \frac{x^{200}}{200} + \frac{x^{199}}{199} + \dots + \frac{x^2}{2} + x + 5$. To find $g'(x)$,we differentiate $g(x)$ wi

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$1$

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(B) Given the function: $g(x) = \frac{x^{200}}{200} + \frac{x^{199}}{199} + \dots + \frac{x^2}{2} + x + 5$. To find $g'(x)$,we differentiate $g(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$: $g'(x) = \frac{200x^{199}}{200} + \frac{199x^{198}}{199} + \dots + \frac{2x}{2} + 1 + 0$. Simplifying the expression,we get: $g'(x) = x^{199} + x^{198} + \dots + x + 1$. Now,substitute $x = 0$ into the derivative: $g'(0) = 0^{199} + 0^{198} + \dots + 0 + 1$. $g'(0) = 1$.

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

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