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For the function \[\] $f(x) = \large\frac{x^{100}}{100}$$+ \large\frac{x^{99}}{99}$$+...+\large\frac{x^2}{2}$$+x+1$ \[\]then $f'(1) is

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The given function is
$f(x) = \large\frac{x^{100}}{100}$$+ \large\frac{x^{99}}{99}$$+...+\large\frac{x^2}{2}$$+x+1$
$ \large\frac{d}{dx}$$f(x) =\large\frac{d}{dx}$$ \bigg[ \large\frac{x^{100}}{100}$$+ \large\frac{x^{99}}{99}$$+...+\large\frac{x^2}{2}$$+x+1 \bigg]$
$ \large\frac{d}{dx}$$f(x) =\large\frac{d}{dx}$$ \bigg( \large\frac{x^{100}}{100}\bigg)$$+\large\frac{d}{dx}$$ \bigg( \large\frac{x^{99}}{99}\bigg)$$+...+\large\frac{d}{dx}$$ \bigg(\large\frac{x^2}{2}\bigg)$$+ \large\frac{d}{dx}$$ (x)+\large\frac{d}{dx}$$ (1)$
On using theorem $\large\frac{d}{dx}$$(x)^n =nx^{n-1}$, we obtain
$\large\frac{d}{dx}$$f(x) \large\frac{100x^{99}}{100}$$+\large\frac{99x^{98}}{99}$$+...+ \large\frac{2x}{2}$$+1+0$
$ = x^{99}+x^{98}+...+x+1$
$ \therefore f'(x) = x^{99}+x^{98}+...+x+1$
At $x=0$
At $x=1$
$f'(1) =1^{99}+1^{98}+...+1+1 = [ 1+1+...+1+1]_{100\: terms} = 1 \times 100 = 100$
Thus, $ f'(1) =100 \times f'(0)$
answered Apr 7, 2014 by thanvigandhi_1

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