Question

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

Answer 1

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{x^{100}}{100}+\frac{x^{99}}{99}+...+\frac{x^2}{2}+x+1\right) \\ =\frac{1}{100}\left(100x^{99}\right)+\frac{1}{99}\left(99x^{98}\right)+...+\frac{1}{2}\left(2x\right)+1+0 \\ =x^{99}+x^{98}+...+x+1 \\ f\text{'}\left(1\right)=1^{99}+1^{98}+...+1+1 \\ =99+1 \\ =100 \\ f\text{'}\left(0\right)=0+0+...+0+1 \\ =1 \\ \mathrm{RHS}=100f\text{'}\left(0\right) \\ =100\left(1\right) \\ =100 \\ =f\text{'}\left(1\right) \\ =\mathrm{LHS} \\ \therefore f\text{'}\left(1\right)=100f\text{'}\left(0\right) \end{gathered}$$