Question

If $f\left(x\right)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\frac{x^{98}}{98}+\dots +x+1,$ show that $f^{\prime }\left(1\right)=100f^{\prime }\left(0\right).$

Answer 1

$f\left(x\right)={\frac{x}{100}}^{100}+{\frac{x}{99}}^{99}+{\frac{x}{98}}^{98}+----+x+1$

Differentiate w.r.t x __

$f^{\prime }\left(x\right)=100\times \frac{1}{100}{x}^{100-1}+\frac{99}{99}{x}^{99-1}+-----1\times x^{1-1}+0.$

$\left\{since\right(x^n{)}^1=\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\}$

$=x^{99}+x^{98}+x^{97}+----x^0$

$f^{\prime }\left(x\right)=x^{99}+x^{98}+x^{97}+-----+.1$

L.H.S. $\Rightarrow f^{\prime }\left(1\right)=1^{99}+1^{98}+----+1=$

= 1+1+ ---- up to 100 terms.

$=100\times 1=100$

R.H.S $\Rightarrow 100f^{\prime }\left(0\right)=100(0^{99}+0^{98}+----+1).$

$=100\times 1=100$

So L.H.S = R.H.S. Hence proved.