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

The given function f(x)=$\frac{x^{100}}{100}$ + $\frac{x^{99}}{99}$ + ... + $\frac{x^2}{2}$ + x + 1
$\frac{d}{dx}$ f(x) = $\frac{d}{dx}$[$\frac{x^{100}}{100}$ + $\frac{x^{99}}{99}$ + ... + $\frac{x^2}{2}$ + x + 1]
$\frac{d}{dx}$f(x) = $\frac{d}{dx}$($\frac{x^{100}}{100}$) + $\frac{d}{dx}$($\frac{x^{99}}{99}$) + ...+ $\frac{d}{dx}$ ($\frac{x^2}{2}$) + $\frac{d}{dx}$(x) + $\frac{d}{dx}$(1)
On using theorem $\frac{d}{dx}$(xn) = nxn-1, we obtain
$\frac{d}{dx}$ f(x) = 1$\frac{100x^{99}}{100}$ + $\frac{99x^{98}}{99}$ + ...+ $\frac{2x}{2}$ + 1+0
= x99 + x98+ ... + x+1
∴f'(x) = x99 + x98+ ... + x+1
At x = 0,
f'(0)=1
At x=1,
ƒ'(1)=199+198+...+1+1=[1+1+...+1+1] 100 terms = 1×100=100
Thus, f'(1)=100× f'(0)