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Have f(x)=㏑
(x+1)/(x+4). Solve P= f'(0)+f'(3)+f'(6)+...+f'(2019).

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Given f(x) = ln(x)

To find P, we need to evaluate the sum of f'(0) + f'(3) + f'(6) + ... + f'(2019).

We know that the derivative of ln(x) is 1/x, so we can write:

f'(x) = 1/x

To find f'(0), we need to use the limit definition of the derivative:

f'(0) = lim (h → 0) [ ln(h) - ln(0) ] / h

Since ln(0) is undefined, we can rewrite this expression as:

f'(0) = lim (h → 0) [ ln(h) ] / h

Using L'Hôpital's rule, we can differentiate the numerator and denominator separately:

f'(0) = lim (h → 0) [ 1/h ] / 1

f'(0) = lim (h → 0) [ 1/h ] = ∞

Note that f'(0) is undefined, since ln(0) is undefined, but the limit approaches infinity.

Similarly, we can find f'(3), f'(6), and so on:

f'(3) = 1/3

f'(6) = 1/6

...

f'(2019) = 1/2019

Now we can substitute these values into the sum:

P = f'(0) + f'(3) + f'(6) + ... + f'(2019)

P = ∞ + 1/3 + 1/6 + ... + 1/2019

The sum on the right-hand side is a finite sum of the form:

1/3 + 1/6 + ... + 1/2019 = ∑(1/3n) where n goes from 1 to 673

This is a telescoping sum, meaning that the terms cancel each other out:

∑(1/3n) = (1/3) + (1/6 + 1/9) + (1/12 + 1/15 + 1/18) + ... + (1/2016 + 1/2019)

= (1/3) + (1/3) + (1/3) + ... + (1/3)

= 673/3

Therefore, we have:

P = ∞ + 673/3

Since f'(0) is undefined, P is also undefined.

User Chris Rasys
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