Derivatives — MATHS STD 11 Science — Question

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

Differentiating with respect to $\text{x},$ We get

$\text{f}'(\text{x})=\text{x}^{99}+\text{x}^{98}+\dots+\text{x}+1+0\dots(\text{i})$

From (i)

f'(1) = 1 + 1 + ...(100 times)

= 100

Again,

f'(0) = 0 + 0 + ... + 1

= 1

Now,

f'(1) = 100 = 100 $\times$ 1 = 100 $\times$ f'(0)

Hence'

f'(1) = 100f'(0)