Evaluating $f'(1)$ for $f(x)=7x-3+\ln(x)$

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In summary, the derivative of a function at a specific point is the slope of the tangent line to the function at that point. To evaluate the derivative, you can use the limit definition or rules of differentiation. For $f(x)=7x-3+\ln(x)$, we can use the sum and power rules to find $f'(1)=8$. The quotient rule cannot be used for finding the derivative in this case. Evaluating $f'(1)$ is important for understanding the behavior of the function near that point and for various applications such as finding maximum or minimum values, rate of change, and graphing.
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karush
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If $f(x)=7x-3+\ln(x),$ then $f'(1)=$
$(A)\, 4\quad (B)\: 5\quad (C)\, 6\quad (D)\,7\quad (E)\,8$
since
$$f'(x)=7+\dfrac{1}{x}$$
so then
$$f'(1)=7+\dfrac{1}{1}=7+1=8\quad (E)$$

ok kinda an easy one but typos and suggestions possible
 
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  • #2
I don't see a question here. If you are asking if your solution is correct, yes it is.
 

FAQ: Evaluating $f'(1)$ for $f(x)=7x-3+\ln(x)$

What does $f'(1)$ represent in this equation?

$f'(1)$ represents the instantaneous rate of change or slope of the function $f(x)$ at the point $x=1$. It gives us the rate at which the function is changing at that specific point.

How do you evaluate $f'(1)$ for $f(x)=7x-3+\ln(x)$?

To evaluate $f'(1)$, we use the power rule for derivatives and the rule for the derivative of natural logarithms. First, we find the derivative of $f(x)$, which is $f'(x)=7+\frac{1}{x}$. Then, we substitute $x=1$ into the derivative, giving us $f'(1)=7+\frac{1}{1}=8$. Therefore, $f'(1)=8$.

Can you explain the significance of $f'(1)$ in this equation?

The value of $f'(1)$ tells us the slope of the function at the specific point $x=1$, which can be interpreted as the rate of change of the function at that point. It also helps us determine the behavior of the function near $x=1$, such as whether it is increasing or decreasing.

Is $f'(1)$ the same as the derivative of $f(x)$?

Yes, $f'(1)$ is the same as the derivative of $f(x)$ at the point $x=1$. The derivative of a function gives us the slope of the function at any given point, and when we substitute a specific value for $x$, we get the derivative at that point, which is $f'(1)$ in this case.

How can we use $f'(1)$ to find the equation of the tangent line to $f(x)$ at $x=1$?

The equation of the tangent line at $x=1$ is given by $y=f'(1)(x-1)+f(1)$. We can use the value of $f'(1)$ that we calculated earlier and the given point $(1,f(1))$ to find the equation of the tangent line. This equation represents a line that is tangent to the curve of $f(x)$ at $x=1$ and has the same slope as the curve at that point.

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