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

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(D) Given the function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$. To find $f'(x)$,we differentiate $f(x)$ with

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(D) Given the function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$. To find $f'(x)$,we differentiate $f(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$: $f'(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \dots + \frac{2x}{2} + 1$. Simplifying this,we get $f'(x) = x^{99} + x^{98} + \dots + x + 1$. Now,substituting $x = 0$ into the derivative: $f'(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1$.

If the function $f(x)$ is defined by $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$,then $f'(0) = $

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