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) = nx^n-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)