How Does Differentiation Determine the Exact Slope of a Curve?

  • Thread starter Thread starter lazyaditya
  • Start date Start date
AI Thread Summary
Differentiation calculates the exact slope of a curve at any given point, as opposed to the slope between two points, which provides an approximation. The derivative, represented as dy/dx, uses limits to find this instantaneous slope. For example, differentiating the equation y = 3x^2 results in dy/dx = 6x, yielding a slope of 18 at x = 3. In contrast, the slope formula (y2-y1)/(x2-x1) gives a value based on two points, which can differ from the derivative's result. Understanding the distinction between these two methods is crucial for accurately analyzing curves in calculus.
lazyaditya
Messages
176
Reaction score
7
We say that a slope can be calculated as the ratio of the perpendicular to the base , Then what does differentiation of the equation shows "dy/dx"?
 
Mathematics news on Phys.org


I'm sorry, what? Can you please clarify your question?
 


We say that a slope can be calculated as the ratio of the perpendicular to the base of a line tangential[/color], Then what does differentiation of the equation shows "dy/dx"?
Differential calculus involves the evaluation of that Δy / Δx but taken to its extreme.
 


I just want to know what actually differentiation is apart from its equation ? what does it mean?
 


Differentiation is a way of calculating the slope at any point along a curve if you know that curve's equation.
 


Ya that's what i m saying suppose y =3x^2 then on differentiation i would get (dy/dx=6x) then at point x=3 the slope would be 18. Then isn't slope also calculated as (y2-y1)/(x2-x1) then at point x = 3 isn't it should be 9 ?
 


lazyaditya said:
Ya that's what i m saying suppose y =3x^2 then on differentiation i would get (dy/dx=6x) then at point x=3 the slope would be 18. Then isn't slope also calculated as (y2-y1)/(x2-x1) then at point x = 3 isn't it should be 9 ?

No, because that equation gives you the slope of the line connecting two points; we don't have two points in this case, we only have one. Remember that when we construct the derivative we actually use the above slope formula to calculate the limit of the slope as the distance of two points approaches zero.
 


lazyaditya said:
Then isn't slope also calculated as (y2-y1)/(x2-x1)
Yes. So if you take two points (x1,y1) and (x2,y2) on that curve, where x1 and x2 are very close to 3, then that quotient will be very close to 18.
 


Then what is the significance of one over the other ? I mean in which situation which one should i prefer ?
 
  • #10


lazyaditya said:
Then what is the significance of one over the other ? I mean in which situation which one should i prefer ?

This has already been explained to you. The slope formula gives you the slope of the line connecting two points; the derivative gives the slope of the tangent line at each point of the function.
 
  • #11


What appears to be puzzling you is the limit process. I recommend you go back and review "limits". It's probably in the chapter of your Calculus book just before the derivative is introduced!
 
  • #12


lazyaditya said:
Then what is the significance of one over the other ? I mean in which situation which one should i prefer ?
Differentiation gives the exact value of the slope at any point you choose. Using the "pair of points" method gives an approximation, you are in effect approximating the curve with a straight line.
 
  • #13


NascentOxygen said:
Differentiation gives the exact value of the slope at any point you choose. Using the "pair of points" method gives an approximation, you are in effect approximating the curve with a straight line.

Thank you i was really confused at this :)
 

Similar threads

Replies
5
Views
3K
Replies
1
Views
257
Replies
5
Views
1K
Replies
8
Views
2K
Replies
22
Views
3K
Replies
53
Views
5K
Replies
6
Views
2K
Replies
9
Views
3K
Back
Top