For f(x)=lnx, construct tables, rounded to four decimals, near x=1, x=2, x=5, and x=10. Use tables to estimate f'(1), f'(2), f'(5) and f'(10). Then guess a general formula for f'(x).

ln (1) = 0

ln(1.1) = .09531
dy/dx = .09531/.1 = .9531

ln 2 = .69315
ln 2.1 = .74194
dy/dx = .04879/.1 = .4879

ln 5 = etc

ln 10 = 2.30258
ln 10.1 = 2.31253
dy/dx = 0.09950 which is about 1/10

looks like 1/x in the end

To construct the tables near the given values for \(x\), we need to evaluate the function \(f(x) = \ln(x)\) at those points. Remember that \(\ln(x)\) represents the natural logarithm of \(x\).

1) For \(x = 1\):
We can calculate \(f(1) = \ln(1) = 0\).

2) For \(x = 2\):
We can calculate \(f(2) = \ln(2) \approx 0.6931\) (rounded to four decimal places).

3) For \(x = 5\):
We can calculate \(f(5) = \ln(5)\approx 1.6094\) (rounded to four decimal places).

4) For \(x = 10\):
We can calculate \(f(10) = \ln(10) \approx 2.3026\) (rounded to four decimal places).

Now, to estimate \(f'(1)\), \(f'(2)\), \(f'(5)\), and \(f'(10)\) (the derivatives of \(f(x)\) at these points), we can use the values from the table.

To estimate the derivative, we can calculate the difference between consecutive function values and divide by the corresponding difference in \(x\):

1) For \(x = 1\):
To estimate \(f'(1)\), we can use the value of \(\ln(2)\) from the table (which is approximately 0.6931). Since the difference in \(x\) is 1, the approximate derivative is \(f'(1) \approx \frac{0.6931 - 0}{2 - 1} = 0.6931\).

2) For \(x = 2\):
To estimate \(f'(2)\), we can determine the difference between \(\ln(5)\) and \(\ln(2)\) (approximately 1.6094 - 0.6931), and divide it by the difference in x, which is 3 (5 - 2). Therefore, the approximate derivative is \(f'(2) \approx \frac{1.6094 - 0.6931}{5 - 2} \approx 0.3054\).

3) For \(x = 5\):
To estimate \(f'(5)\), we can determine the difference between \(\ln(10)\) and \(\ln(5)\) (approximately 2.3026 - 1.6094), and divide it by the difference in x, which is 5 (10 - 5). Therefore, the approximate derivative is \(f'(5) \approx \frac{2.3026 - 1.6094}{10 - 5} \approx 0.1786\).

4) For \(x = 10\):
To estimate \(f'(10)\), we can estimate the difference between \(\ln(10)\) and \(\ln(5)\) (approximately 2.3026 - 1.6094), and divide it by the difference in x, which is 5 (10 - 5). Therefore, the approximate derivative is \(f'(10) \approx \frac{2.3026 - 1.6094}{10 - 5} \approx 0.1786\).

Now let's try to guess a general formula for \(f'(x)\) based on the table values:
From the table, we observe that the approximate derivative appears to be roughly constant and equal to 0.6931. Therefore, we guess that the general formula for \(f'(x)\) is \(f'(x) = \frac{1}{x}\).

Please keep in mind that this is just a guess based on our observations from the table and needs to be further explored and verified through mathematical techniques like differentiation or limit calculations.