Question

If f(x) = (x+1) (x+2) (x+3).......(x+n) , then find f'(0).

Answer 1

Hey there!!!

Here, f(x) = (x+1) (x+2) (x+3)......(x+n)

log f(x) = log { (x+1) (x+2) (x+3) ........(x+n) }

Differentiating both sides with respect to x, we get:

1/f(x) f'(x) = 1/(x+1) +1(x+2) + 1/(x+3) +..........+ 1/(x+n)
Therefore,
f'(0) = f(0) { 1 + 1/2 + 1/3 + .........+1/n }
        = (1.2.3. ....... .n) {1 + 1/2 + 1/3 + ....... 1/n }
        = (n)! { 1 + 1/2 + 1/3 + .........1/n }



HOPE IT HELPS!!!!

# RUHANIKA  

Answer 2

Hope this may help you
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