For the function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$,prove that $f^{\prime}(1) = 100 f^{\prime}(0)$.

(N/A) The given function is $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$.
Taking the derivative with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx} \left( \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1 \right)$.
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f^{\prime}(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \dots + \frac{2x}{2} + 1 + 0$.
$f^{\prime}(x) = x^{99} + x^{98} + \dots + x + 1$.
At $x = 0$,$f^{\prime}(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1$.
At $x = 1$,$f^{\prime}(1) = 1^{99} + 1^{98} + \dots + 1 + 1$.
Since there are $100$ terms in the sum,$f^{\prime}(1) = 1 \times 100 = 100$.
Thus,$f^{\prime}(1) = 100 \times 1 = 100 f^{\prime}(0)$.
Hence,$f^{\prime}(1) = 100 f^{\prime}(0)$ is proved.