Question

If $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\frac{x^{98}}{98}$ +...+x+1$,showthat$f'(1)=100 f'(0)$.

Answer 1

Differentiate the function $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\frac{x^{98}}{98}+\cdots +x+1$ with respect to $x$.

$f^{\prime }\left(x\right)=x^{99}+x^{98}+x^{97}+\cdots +1$

$f^{\prime }\left(0\right)={\left(0\right)}^{99}+{\left(0\right)}^{98}+{\left(0\right)}^{97}+\cdots +1$

$=1$

$f^{\prime }\left(1\right)={\left(1\right)}^{99}+{\left(1\right)}^{98}+{\left(1\right)}^{97}+\cdots +1$

$=1+1+\cdots \left(100times\right)$

$=100$

$f^{\prime }\left(1\right)=100\times 1$

$=100f^{\prime }\left(0\right)$

Hence proved.