Mastering Derivatives: Tips and Tricks for Calculus Success

  • Thread starter Thread starter mathzeroh
  • Start date Start date
  • Tags Tags
    Derivatives
Click For Summary
SUMMARY

The discussion centers on understanding derivatives in calculus, specifically how to find the derivative at a point using a graph. Participants clarify that the derivative at a specific point, such as f'(-4), represents the slope of the tangent line at that point, not the equation of the tangent line itself. The conversation emphasizes the importance of recognizing where the graph is linear to determine the slope accurately. The final conclusion confirms that the derivative at f'(-4) is -2, as derived from the slope calculation between two points on the graph.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly derivatives.
  • Familiarity with graph interpretation and slope calculation.
  • Knowledge of tangent lines and their significance in calculus.
  • Ability to apply the concept of derivatives to specific points on a graph.
NEXT STEPS
  • Study the definition and properties of derivatives in calculus.
  • Learn how to calculate slopes of tangent lines from graphs.
  • Explore the concept of differentiability and where derivatives do not exist.
  • Practice using the quotient rule for differentiation in various functions.
USEFUL FOR

Students struggling with calculus, particularly those learning about derivatives and their applications. This discussion is beneficial for anyone seeking to improve their understanding of how to analyze graphs in relation to calculus concepts.

mathzeroh
Messages
98
Reaction score
0
In light of my current failure of the first exam of my calculus class, I've been a bit discouraged, but i try not to let these grades get me down.

with that said, can anyone be so kind as to help me wrap my mind around the concept of derivatives? i have a shaky understanding of them as it is, therefore, any help or tips that you all may have, is greatly appreciated. how did you learn about them? what helped you go, "OH! i get it now!" ?

for example, how do i find this:

f'(-4)

and all I'm given is a graph of f(x).

any help on this??:confused: i greatly appreciate it in advance! thank you!

i tried my best to duplicate the graph, though it may be a poor representation. my apologies.
 

Attachments

  • Graph of f(x).JPG
    Graph of f(x).JPG
    6.3 KB · Views: 452
Physics news on Phys.org
Well think about what a derivative is? What does the derivative of a function at a specific point mean?
 
i know that at a specific point, it's the equation of the tangent line.

but what about that example problem?:confused:
 
mathzeroh said:
i know that at a specific point, it's the equation of the tangent line.

but what about that example problem?:confused:

Well, that's your first problem- you know wrong!

The derivative of a function, at a specific point, is the slope of the tangent line to the graph at that point, a number, not the equation of the tangent line.

Now, it isn't easy to sketch the graph of f ' just by eyeballing the graph of f but in your specific case it looks like the graph is largely made of straight lines. Can you find the slope of those? Of course, where the graph is a straight line, the derivative at each point is the slope. Also you should be able to easily see where the derivative is positive, negative, or zero.

Be careful about places where the derivative does not exist.
 
HallsofIvy said:
Now, it isn't easy to sketch the graph of f ' just by eyeballing the graph of f but in your specific case it looks like the graph is largely made of straight lines. Can you find the slope of those? Of course, where the graph is a straight line, the derivative at each point is the slope. Also you should be able to easily see where the derivative is positive, negative, or zero.

Be careful about places where the derivative does not exist.

i don't understand. what about the quotient formulas?
 
mathzeroh said:
i don't understand. what about the quotient formulas?

I don't understand what that has to do with you're initial problem... You aren't differentiating a function you're trying to estimate the derivative at a point.
 
The derivative at a point is the slope of the tangent line at that point. If the graph happens to be a straight line, then the derivative at each point is the slope of that line.

You asked specifically about f'(-4). From your graph it looks like the graph there is a straight line through (-3, -2) and (-5, 2). What is the slope of that line?
 
the slope of that line is:

2-(-2)/(-5)-(-3)

=2+2/-5+3

=4/-2

=(-2)

is that it? :rolleyes: :confused:
 
HallsofIvy said:
The derivative at a point is the slope of the tangent line at that point. If the graph happens to be a straight line, then the derivative at each point is the slope of that line.

You asked specifically about f'(-4). From your graph it looks like the graph there is a straight line through (-3, -2) and (-5, 2). What is the slope of that line?

so was that it? -2?
 
  • #10
mathzeroh said:
so was that it? -2?
Is it what? You were originally asking about f '(-4). One of the first things you should have learned about the derivative at a point is that it is the slope of the tangent line at that point. In particular if a graph is a straight line on an interval, its derivative at any point in that interval is its slope.
 
  • #11
HallsofIvy said:
Is it what? You were originally asking about f '(-4). One of the first things you should have learned about the derivative at a point is that it is the slope of the tangent line at that point. In particular if a graph is a straight line on an interval, its derivative at any point in that interval is its slope.
But I thought his post #8, apart from some missing parentheses, is correct... :approve:
Isn't it?
 
  • #12
HallsofIvy said:
Is it what? You were originally asking about f '(-4). One of the first things you should have learned about the derivative at a point is that it is the slope of the tangent line at that point. In particular if a graph is a straight line on an interval, its derivative at any point in that interval is its slope.

well thanks a lot for your help. that makes sense. i guess its just something that will take a little while to sink in completely, but i know that once know it fully, i'll be better at knowing what to do. so the answer to my question of f'(-4) in that problem was -2, though right?:blushing: (just like VietDao said?)
 

Similar threads

Replies
16
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
2
Views
1K
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
12
Views
2K
Replies
10
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K
Replies
3
Views
2K