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

Here f(x) = x^ 100 100 + x^ 99 99 + ... + x^2 2 + x + 1 f ' (x) = d dx [ x^ 100 100 + x^ 99 99 + ... + x^2 2 + x + 1 ] = 1 100 d dx ( x^ 100 ) + 1 99 d dx ( x^ 99 ) + ... + 1 2 d dx ( x^2 ) + d dx (x) + d dx (1) = 1 100 100 x^ 99 + 1 99 99 x^ 98 + ... + 1 2 2x + 1 + 0 = x99 + x98 + ... + x + 1 Now f'(1) = (1)99 + (1)98 + ...+(1) + 1 = 100 f'(0) = (0)99 + (0)98 + ...+ 0 + 1 = 1 Which shows that f'(1) = 100f'(0)

For the function f(x) = x^ 100 100 + x^ 99 99 + ... + x^2 2 + x +…

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Question

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

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Answer

Here $f(x) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1$
$f{\text{'}}(x) = \frac{d}{{dx}}\left[ {\frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + ... + \frac{{{x^2}}}{2} + x + 1} \right]$
$= \frac{1}{{100}}\frac{d}{{dx}}({x^{100}}) + \frac{1}{{99}}\frac{d}{{dx}}({x^{99}}) + ... + \frac{1}{2}\frac{d}{{dx}}({x^2}) + \frac{d}{{dx}}(x) + \frac{d}{{dx}}(1)$
$= \frac{1}{{100}} \times 100{x^{99}} + \frac{1}{{99}} \times 99{x^{98}} + ... + \frac{1}{2} \times 2x + 1 + 0$
= x⁹⁹ + x⁹⁸ + ... + x + 1
Now f'(1) = (1)⁹⁹ + (1)⁹⁸ + ...+(1) + 1 = 100
f'(0) = (0)⁹⁹ + (0)⁹⁸ + ...+ 0 + 1 = 1
Which shows that f'(1) = 100f'(0)

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