Find derivative of function given the values

In summary, the author attempted to approximate the value of erf '(1) by taking the average from both sides of erf(.01). However, their estimate is too large by a factor of 2.
  • #1
Painguy
120
0

Homework Statement


You are given the following values for the errror function f HxL = erf HxL.
erf (0)= 0
erf (1) = 0.84270079
erf(.1)= 0.11246292
erf(.01) = 0.01128342
Use this information to approximate the value of erf '(1). Explain how you arrived at your answer.

Homework Equations





The Attempt at a Solution


I basically took the average from both sides of erf(.01)
(((erf(0)-erf(.01))/.01)+((erf(.01)-erf(.1))/(-.09)))/2 = (1.128342 + 1.124216667)/2 =1.126279335

I ended up with 1.126279335 so I'm assuming its something close to that like 1.12? Am I right or wrong?
 
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  • #2
Painguy said:

Homework Statement


You are given the following values for the errror function f HxL = erf HxL.
erf (0)= 0
erf (1) = 0.84270079
erf(.1)= 0.11246292
erf(.01) = 0.01128342
Use this information to approximate the value of erf '(1). Explain how you arrived at your answer.

Homework Equations





The Attempt at a Solution


I basically took the average from both sides of erf(.01)
(((erf(0)-erf(.01))/.01)+((erf(.01)-erf(.1))/(-.09)))/2 = (1.128342 + 1.124216667)/2 =1.126279335

I ended up with 1.126279335 so I'm assuming its something close to that like 1.12? Am I right or wrong?

Why would you use values near 0 when you are interested in erf'(1)? Did you type the last two data points correctly?
 
  • #3
Painguy said:

Homework Statement


You are given the following values for the errror function f HxL = erf HxL.
What is the HxL business?
Painguy said:
erf (0)= 0
erf (1) = 0.84270079
erf(.1)= 0.11246292
erf(.01) = 0.01128342
Use this information to approximate the value of erf '(1). Explain how you arrived at your answer.

Homework Equations





The Attempt at a Solution


I basically took the average from both sides of erf(.01)
(((erf(0)-erf(.01))/.01)+((erf(.01)-erf(.1))/(-.09)))/2 = (1.128342 + 1.124216667)/2 =1.126279335
I can't see that this will do you much good. To get an estimate of the derivative at x = 1, you need values that are reasonably close to x = 1. The two that are closest are x = .1 and x = 1.
Painguy said:
I ended up with 1.126279335 so I'm assuming its something close to that like 1.12? Am I right or wrong?
Your estimate is too large by at least a factor of 2.
 
  • #4
Sorry i meant to say at erf'(0).that was a typo on my part
 

1. What is 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 can be interpreted as the instantaneous rate of change of the function.

2. How do you find the derivative of a function?

The derivative of a function can be found using the process of differentiation. This involves using a set of rules and formulas to manipulate the function and find its rate of change at a specific point. It requires knowledge of algebra, trigonometry, and calculus.

3. Can the derivative of a function be negative?

Yes, the derivative of a function can be negative. This indicates that the function is decreasing at that point, meaning its rate of change is negative. A positive derivative would indicate that the function is increasing at that point.

4. What are the applications of finding the derivative of a function?

The derivative of a function has many real-world applications, including in physics, engineering, economics, and statistics. It can be used to calculate velocity, acceleration, optimization problems, and more.

5. Can the derivative of a function be undefined?

Yes, the derivative of a function can be undefined at certain points. This occurs when the function is not continuous or differentiable at that point, or when there is a vertical tangent line. In these cases, the derivative does not exist.

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