Differentiate

Calculus Level 3

If f ( x ) = ∏ n = 1 100 ( x − n ) n ( 101 − n ) f(x)= \displaystyle\prod\limits_{n=1}^{100} (x-n)^{n(101-n)} , then find f ( 101 ) f ′ ( 101 ) \dfrac{f(101)}{f'(101)} .

1 6570 \dfrac1{6570} 1 5050 \dfrac1{5050} 1 4500 \dfrac1{4500} 1 100 \dfrac1{100}

Prithwish Roy by Prithwish Roy , 20, India

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1 solution

Brian Moehring
Mar 25, 2017

Note that f ′ ( 101 ) f ( 101 ) = d d x [ ln ⁡ f ( x ) ] ∣ x = 101 = d d x [ ∑ n = 1 100 n ( 101 − n ) ln ⁡ ( x − n ) ∣ x = 101 = ∑ n = 1 100 n ( 101 − n ) 101 − n = ∑ n = 1 100 n = 100 ( 100 + 1 ) 2 = 5050 \begin{aligned} \frac{f'(101)}{f(101)} &= \frac{d}{dx}\left[\ln f(x)\right]\Big|_{x=101} \\ &= \frac{d}{dx}\left[ \sum_{n=1}^{100} n(101-n)\ln(x-n)\right|_{x=101} \\ &= \sum_{n=1}^{100} \frac{n(101-n)}{101-n} \\ &= \sum_{n=1}^{100} n = \frac{100(100+1)}{2} = 5050 \end{aligned}

Therefore, f ( 101 ) f ′ ( 101 ) = 1 5050 \frac{f(101)}{f'(101)} = \frac{1}{5050}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Wouldntit be continued product instead of summation in you second step

Md Zuhair - 4 years, 2 months ago

Okay now I got it... it is ln at front so summation

Md Zuhair - 4 years, 2 months ago

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