Question

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

Answer 1

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

f'(x) = $\frac{d}{dx}[\frac{x^{100}}{100}+\frac{x^{99}}{99}+...+\frac{x^2}{2}+x+1]$

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

= $\frac{1}{100}\times 100x^{99}+\frac{1}{99}\times 99x^{98}+...+\frac{1}{2}\times 2x+1+0$

Now f'(1) = $(1{)}^{99}+(1{)}^{98}+...+\left(1\right)+1=100$
$f^{\prime }\left(0\right)=(0{)}^{99}+(0{)}^{98}$+...+$0+1=1$

which shows that f'(1) = 100 f'(0).