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

• MHB
• karush
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.
karush
Gold Member
MHB
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

I don't see a question here. If you are asking if your solution is correct, yes it is.

## 1. 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.

## 2. 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, $f'(1)=8$.

## 3. 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.

## 4. 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.

## 5. 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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