Derivatives.

Calculus Level 2

If f ( x ) = 1 + x + x 2 + x 3 + ⋯ + x 2017 + x 2018 f(x)=1+x+x^2+x^3+\dots+x^{2017}+x^{2018} then what is f ′ ( 1 ) = ? f'(1)=?

2019 2 \frac{2019}{2} ( 2018 ) ( 2019 ) 2 \frac{(2018)(2019)}{2} 2018 2 \frac{2018}{2} 201 8 2 2 \frac{2018^2}{2} ( 2017 ) ( 2018 ) 2 \frac{(2017)(2018)}{2}

Hana Wehbi by Hana Wehbi , 51, USA

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2 solutions

Ram Mohith
Oct 2, 2018

Given, f ( x ) = 1 + x + x 2 + x 3 + . . . + x 2017 + x 2018 f(x) = 1 + x + x^2 + x^3 + ... + x^{2017} + x^{2018} On differentiating with respect to x x , we get : f ′ ( x ) = 0 + 1 + 2 x + 3 x 2 + . . . + 2017 x 2016 + 2018 x 2017 f'(x) = 0 + 1 + 2x + 3x^2 + ... + 2017x^{2016} + 2018x^{2017} On substituting x = 1 x = 1 , we get f ′ ( 1 ) = 1 + 2 + 3 + . . . + 2017 + 2018 f'(1) = 1 + 2 + 3 + ... + 2017 + 2018 ⟹ f ′ ( 1 ) = ( 2018 ) ( 2019 ) 2 \implies f'(1) = \dfrac{(2018)(2019)}{2}

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Krishna Karthik
Oct 6, 2018

d d x \frac{d}{dx} ( 1 + x + x 2 + x 3 . . . 1+x+x^2+x^3... ) = till x power 2018.

1 + 2 x + 3 x 2 + 4 x 3 . . . 1+2x+3x^2+4x^3... till the last term. substituting 1 into this, 1 power anything is just one, so f prime of 1 is just 1 + 2 + 3 + 4 + 5... + 2017 + 2018 1+2+3+4+5...+2017+2018 .

The formula for this sum is obviously n ( n + 1 ) 2 \frac{n(n+1)}{2} in which n is the last term. therefore the first option is the answer.

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