For the functionf(x)= x^ 100 100 + x^ 99 99 +...+ x^2 2 +x+1. Prove that f'(1)=100f'(0).

The given function is f(x)= x^ 100 100 + x^ 99 99 +...+ x^2 2 +x+1 d dx f(x)= d dx [ x^ 100 100 + x^ 99 99 +...+ x^2 2 +x+1 ] d dx f(x)= d dx ( x^ 100 100 )+ d dx ( x^ 99 99 )+...+ d dx ( x^2 2 )+ d dx (x )+ d dx (1 ) On using theorem d dx (x)^n=nx^ n-1 , we obtain d dx f(x)= 100x^ 99 100 + 99x^ 98 99 +...+ 2x 2 +1+0 =x^ 99 +x^ 98 +...+x+1 '(x)=x^ 99 +x^ 98 +...+x+1 At x=0, f'(0)=1 At x=1,f'(1)=1^ 99 +1^ 98 +...+1+1=[1+1+...+1+1]_ 100terms =1 100=100 Thus,f'(1)=100 '(0)

For the functionf(x)= x^ 100 100 + x^ 99 99 +...+ x^2 2 +x+1. Pro…

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Question

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

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Answer

The given function is $\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1$ $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{\text{d}}{\text{dx}}\bigg[\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1\bigg]$ $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^{100}}{100}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^{99}}{99}\bigg)+...+\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^2}{2}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(\text{x}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(1\bigg)$ On using theorem $\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}=\text{nx}^{\text{n}-1}$, we obtain $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{100\text{x}^{99}}{100}+\frac{99\text{x}^{98}}{99}+...+\frac{2\text{x}}{2}+1+0$ $=\text{x}^{99}+\text{x}^{98}+...+\text{x}+1$ $\therefore\text{f}'(\text{x})=\text{x}^{99}+\text{x}^{98}+...+\text{x}+1$ At $\text{x}=0$, $\text{f}'(0)=1$ At $\text{x}=1$,$\text{f}'(1)=1^{99}+1^{98}+...+1+1=[1+1+...+1+1]_{100\text{terms}}=1\times100=100$ Thus,$\text{f}'(1)=100\times\text{f}'(0)$