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Question
For the function
f(x) = `x^100/100 + x^99/99 + ...+ x^2/2 + x + 1`
Prove that f'(1) = 100 f'(0)
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Solution
The given function is
`f(x) = x^100/100 + x^99/99 + ....... + x^2/2 + x + 1`
∴ `d/(dx) f(x) = [(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 + (99^98)/99 + ... + (2x)/2 + 1 + 0`
= x99 + x98 + ..... + x + 1
∴ f'(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]_(100 "terms")` = 1 × 100 = 100
Thus, f'(1) = 100 × f'(0)
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