If f(x) = 7x – 3 + ln x, then f'(1) =
To find f'(1), we need to find the derivative of the function f(x) with respect to x and evaluate it at x = 1
To find f'(1), we need to find the derivative of the function f(x) with respect to x and evaluate it at x = 1.
The derivative of f(x) can be found by applying the rules of differentiation. Let’s break down f(x) into its individual components:
f(x) = 7x – 3 + ln(x)
The first part, 7x, is a simple linear function, and the derivative of a linear function is simply the coefficient of x:
d/dx (7x) = 7
The second part, -3, is a constant, and the derivative of a constant is always 0:
d/dx (-3) = 0
The last part, ln(x), involves a natural logarithm. The derivative of ln(x) is 1/x:
d/dx (ln(x)) = 1/x
Now, we can add up the derivatives of the individual components:
f'(x) = d/dx (7x) – d/dx (3) + d/dx (ln(x))
= 7 – 0 + 1/x
= 7 + 1/x
To find f'(1), we substitute x = 1 into the derivative expression:
f'(1) = 7 + 1/1
= 7 + 1
= 8
Therefore, f'(1) = 8.
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