How can I calculate the derivative of this function?

So, you want to find the slope of f at the points where the graph of f crosses the vertical lines x=1, x=2, x=4, x=7.In summary, the conversation discusses determining the values of f'(a) for a given function, using the definition of the derivative and the geometric interpretation of the derivative as the slope or instantaneous rate of change of a function at a given point. The values of f'(a) can be easily determined by visually inspecting the slope of the function at the points where it crosses the vertical lines x=1, x=2, x=4, x=7.
  • #1
GaussianSurface

Homework Statement


Let f(x) be the function whose graph is shown below (I'll upload the image)

Determine f'(a) for a = 1,2,4,7.

f'(1) =

f'(2) =

f'(4) =

f'(7) =

Use one decimal.

Homework Equations


f(x+h)-f(x)/h

The Attempt at a Solution


Hi everybody I was trying to do this function but I haven't been able to do it even though I'm supposing some things about how this problem can be compute, so I want to ask for help so some of you can guide me to the final result, hope some of you all can help me.
 

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  • #2
When consdering some function ##f(x)##, what is a geometric interpretation of ##f'(a)##?
 
  • #3
MarkFL said:
When consdering some function ##f(x)##, what is a geometric interpretation of ##f'(a)##?

I guess it's not specified, actually I had the same problem I was wondering what's ƒ'(a) since it is not in the graph they gave me...
 
  • #4
What I mean is, in your study of the derivative, what have you been taught that the derivative can represent geometrically, that is, in relation to the graph of the function?
 
  • #5
MarkFL said:
What I mean is, in your study of the derivative, what have you been taught that the derivative can represent geometrically, that is, in relation to the graph of the function?
Well, it can represent the slope or the instantaneuos rate of change. that's how they've taught me, is that what you wanted to know?
 
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  • #6
GaussianSurface said:
Well, it can represent the slope or the instantaneuos rate of change. that's how they've taught me, is that what you wanted to know?

Yes, that's what I was after. :biggrin:

So, can you determine the instantaneous slope of the given function's graph at the given values of ##a##?
 
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  • #7
MarkFL said:
Yes, that's what I was after. :biggrin:

So, can you determine the instantaneous slope of the given function's graph at the given values of ##a##?

Let me see if I got you.

Since you asked about to determine the insatataneous slope I'd say that we can do it by the defnition of derivative ( the one that I posted above ) so far am I going in the good way?
 
  • #8
GaussianSurface said:
Let me see if I got you.

Since you asked about to determine the insatataneous slope I'd say that we can do it by the defnition of derivative ( the one that I posted above ) so far am I going in the good way?

It's actually more simple than that...you can "eyeball" the slope of the function at the given points...:wink:
 
  • #9
MarkFL said:
It's actually more simple than that...you can "eyeball" the slope of the function at the given points...:wink:
Mmmm... Shall I take the prime points at x and compare with f'(a) and so on...?
 
  • #10
Place a dot on your curve above each of the x-values where you are asked to find a value for f’

This will highlight exactly which points on the plot are under discussion.
 
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  • #11
GaussianSurface said:
Mmmm... Shall I take the prime points at x and compare with f'(a) and so on...?

First, let's consider the function where ##0<x<3##...we have a horizontal line...what is the slope of a horizontal line?
 
  • #12
NascentOxygen said:
Place a dot on your curve above each of the x-values where you are asked to find a value for f’

This will highlight exactly which points on the plot are under discussion.
Ohh! I think got it when the prime values get x then a value f'(x) changes with it, doesn't it?
 
  • #13
GaussianSurface said:
Ohh! I think got it when the prime values get x then a value f'(x) changes with it, doesn't it?
I can't decide whether you can see the answers or not. As MarkFL indicated, the values of f' in this exercise can be decided by mere inspection. Remember, f'(b) is the slope of the curve f around point b.
 

Related to How can I calculate the derivative of this function?

1. What is the definition of a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is the slope of the tangent line at that point and is calculated as the limit of the ratio of the change in the output of the function to the change in the input as the change in input approaches zero.

2. How do I find the derivative of a function using the power rule?

The power rule states that the derivative of a function raised to a constant power is equal to the constant times the function raised to the power minus one. For example, if the function is f(x) = x^3, the derivative would be f'(x) = 3x^2.

3. Can I use the product rule to find the derivative of a product of two functions?

Yes, the product rule states that the derivative of a product of two functions f(x) and g(x) is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. In other words, (f(x)g(x))' = f(x)g'(x) + g(x)f'(x).

4. How do I find the derivative of a function using the quotient rule?

The quotient rule states that the derivative of a quotient of two functions f(x) and g(x) is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. In other words, (f(x)/g(x))' = (g(x)f'(x) - f(x)g'(x)) / g(x)^2.

5. Can I use the chain rule to find the derivative of a composite function?

Yes, the chain rule allows us to find the derivative of a composite function by taking the derivative of the outer function and multiplying it by the derivative of the inner function. In other words, if f(x) = g(h(x)), then f'(x) = g'(h(x)) * h'(x).

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