For the Function F ( X ) = X 100 100 + X 99 99 + . . . + X 2 2 + X + 1 . - Mathematics

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

Solution

$f'\left( x \right) = \frac{d}{dx}\left( \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . + \frac{x^2}{2} + x + 1 \right)$
$= \frac{1}{100}\left( 100 x^{99} \right) + \frac{1}{99}\left( 99 x^{98} \right) + . . . + \frac{1}{2}\left( 2x \right) + 1 + 0$
$= x^{99} + x^{98} + . . . + x + 1$
$f'\left( 1 \right) = 1^{99} + 1^{98} + . . . + 1 + 1$
$= 99 + 1$
$= 100$
$f'\left( 0 \right) = 0 + 0 + . . . + 0 + 1$
$= 1$
$RHS = 100 f'\left( 0 \right)$
$= 100\left( 1 \right)$
$= 100$
$= f'\left( 1 \right)$
$= LHS$
$\therefore f'\left( 1 \right) = 100 f'\left( 0 \right)$

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.3 | Q 26 | Page 34