For the function
Question:

For the function

$f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1$

Prove that $f^{\prime}(1)=100 f^{\prime}(0)$

Solution:

The given function is

$f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1$

$\frac{d}{d x} f(x)=\frac{d}{d x}\left[\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1\right]$

$\frac{d}{d x} f(x)=\frac{d}{d x}\left(\frac{x^{100}}{100}\right)+\frac{d}{d x}\left(\frac{x^{99}}{99}\right)+\ldots+\frac{d}{d x}\left(\frac{x^{2}}{2}\right)+\frac{d}{d x}(x)+\frac{d}{d x}(1)$

On using theorem $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$, we obtain

$\frac{d}{d x} f(x)=\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\ldots+\frac{2 x}{2}+1+0$

$=x^{99}+x^{98}+\ldots+x+1$

$\therefore f^{\prime}(x)=x^{99}+x^{98}+\ldots+x+1$

At $x=0$

$f^{\prime}(0)=1$

$f^{\prime}(1)=1^{99}+1^{98}+\ldots+1+1=[1+1+\ldots+1+1]_{100 \mathrm{terms}}=1 \times 100=100$

Thus, $f^{\prime}(1)=100 \times f^{\prime}(0)$

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