How Do You Find F'(8) Using Given Function Values?

[Dantes]

If I get a question that gives me a function defined in variables and then certain F(X) and F'(X) with numbers that equal the slopes, and I need to find the derivative of the first function do I just substitute in.

For example:

Find the value of F'(8) when

$$f(x)/f(x)-g(x)$$

while $$f(8)= 3, f'(8) = 2 , g(8) = 4 , g'(8) = 3$$

do I just do:

$$F(X) = 3 / 3-4$$ and then use Newtonian qoutient to got the derivative of F(X) at 8 ?

 

[maverick280857]

The question you have asked Dantes is part of functional equations. Normally, one tries to use all the data given in a problem and ingeniously get to an equation/relationship that was sought at the start.

According to your problem, you have to find the derivative of F at the point x = 8.

First off, do you mean that

$$F(x) = f(x)/f(x)-g(x)$$
?

Secondly, how do you think you can find the derivative at a point after substituting the value of the independent variable in the function? Essentially if you have to find f'(c)--the derivative of f(x) at a point c--you would (even by first principles) find first f'(x) at a general point x in the domain of f(x) where f(x) is derivable. Then, you would substitute x = c to get f'(c). However, if you find f(c) first, and differentiate, you get a zero as f(c) is a constant value. This point, however has nothing explicit to do with your problem except the last line of your post as I comprehend it.

Let's say you have

$$F(x) = f(x)/f(x)-g(x)$$

Then for all f(x) not equal to zero

$$F(x) = 1-g(x)$$

so that

$$\frac{dF}{dx} = -\frac{dg}{dx} = -g'(x)$$

Cheers
Vivek

 

[M98Ranger]

Yes, you are correct. To find the derivative of the first function, you can use the Newtonian quotient by substituting the given values for f(8), f'(8), g(8), and g'(8). This will give you the value of F'(8). However, it is important to note that the quotient rule is used when the function is in the form of f(x)/g(x), so make sure to rearrange the given function in that form before substituting the values.