Differential (derivative from first principles)

Instead of doing this, you can try to simplify the above expression in a way that it is obvious that the limit is as you want it.There are different ways to prove the limit. One way is to use the definition of the derivative:$$\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=f'(a).$$So, we want to calculate $$\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}=\lim_{h\to 0} \frac{\frac{1}{1-(a+h)}-\frac{1}{1-a}}{h}.$$We can simplify this expression using the common denominator (
  • #1
5ymmetrica1
88
0
1. Homework Statement [/b]
find from the first principles the derivitive of f(x) = [itex]\frac{1}{1 - x}[/itex]

Homework Equations


f'(a) = [itex]\frac{f(a+h) - f(a)}{h}[/itex]


The Attempt at a Solution



f'(x) = [itex]\frac{\frac{1}{1-(x+h)} - \frac{1}{1-x}}{h}[/itex]

f'(x) = [itex]\frac{(1-x) - (1-(x+h))}{h(1-x)(1-(x+h))}[/itex]

but I'm unsure where to go from here
 
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  • #2
5ymmetrica1 said:
1. Homework Statement [/b]
find from the first principles the derivitive of f(x) = [itex]\frac{1}{1 - x}[/itex]

Homework Equations


f'(a) = [itex]\frac{f(a+h) - f(a)}{h}[/itex]


The Attempt at a Solution



f'(x) = [itex]\frac{\frac{1}{1-(x+h)} - \frac{1}{1-x}}{h}[/itex]

f'(x) = [itex]\frac{(1-x) - (1-(x+h)}{h(1-x)(1-(x+h)}[/itex]

but I'm unsure where to go from here
You miss some parentheses. After fixing it, expand the expressions and simplify. Then take the limit h-->0.


ehild
 
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  • #3
by opening up the top parentheses and expanding the bottom I get

[itex]\frac{1-x-1-x+h}{h(1-3x+x^2)}[/itex]

and upon simplifying I have...

[itex]\frac{-2x+h}{h-3xh+x^2h}[/itex]

is there correct or am I doing it wrong?
 
  • #4
5ymmetrica1 said:
by opening up the top parentheses and expanding the bottom I get

[itex]\frac{1-x-1-x+h}{h(1-3x+x^2)}[/itex]

and upon simplifying I have...

[itex]\frac{-2x+h}{h-3xh+x^2h}[/itex]

is there correct or am I doing it wrong?

You are doing it wrong. -(1-x+h)=(-1+x-h).
 
  • #5
5ymmetrica1 said:
f'(a) = [itex]\frac{f(a+h) - f(a)}{h}[/itex]
The equality sign tells us that the expression on the left represents the same number as the expression on the right. So this statement isn't true.

5ymmetrica1 said:
I'm unsure where to go from here
As Dick has already said, it's mainly a matter of simplifying the result correctly before you take the limit h→0.
 
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  • #6
Fredrik said:
The equality sign tells us that the expression on the left represents the same number as the expression on the right. So this statement isn't true.
Do you mean there is something wrong with the equation?

This is exactly how it is written in my textbook
f'(a) = [itex]lim_{h→0}[/itex][itex]\frac{f(a+h) - f(a)}{h}[/itex]

accept h→0 is directly underneathe the lim part, but I am not sure how to write that on PF.
 
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  • #7
f'(x) = [itex](\frac{h}{(1-x-h)(1-x)})(\frac{1}{h})[/itex]

f'(x) = [itex]\frac{1}{(1-x-h)(1-x)}[/itex]

f'(x) = [itex]\frac{1}{(1-x)^2}[/itex]

checking

f(x) = (1-x)-1

f'(x) = -(1-x)-2(-1)

f'(x) = [itex]\frac{1}{(1-x)^2}[/itex]

is this correct?
 
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  • #8
5ymmetrica1 said:
Do you mean there is something wrong with the equation?

This is exactly how it is written in my textbook
f'(a) = [itex]lim_{h→0}[/itex][itex]\frac{f(a+h) - f(a)}{h}[/itex]

accept h→0 is directly underneathe the lim part, but I am not sure how to write that on PF.
I meant that you didn't include the "lim" part of it at all. You just wrote that f'(a) is equal to (f(a+h)-f(a))/h. This would mean that both notations represent the same number, and that there's no need to take a limit.

Hit the quote button to see how I'm doing this.
$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$ See the LaTeX FAQ for more.
 
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  • #9
5ymmetrica1 said:
f'(x) = [itex](\frac{h}{(1-x-h)(1-x)})(\frac{1}{h})[/itex]

f'(x) = [itex]\frac{1}{(1-x-h)(1-x)}[/itex]

f'(x) = [itex]\frac{1}{(1-x)^2}[/itex]

checking

f(x) = (1-x)-1

f'(x) = -(1-x)-2(-1)

f'(x) = [itex]\frac{1}{(1-x)^2}[/itex]

is this correct?
You shouldn't include the part "f'(x)=" unless you also include a ##\lim_{h\to 0}##. You can start this way:
$$
\frac{f(x+h)-f(x)}{h}=\frac{\frac{1}{1-x-h}-\frac{1}{1-x}}{h} =\frac{\frac{1-x-(1-x-h)}{(1-x-h)(1-x)}}{h} =\frac{1}{(1-x-h)(1-x)}
$$ Now you have to prove that regardless of the value of x, the right-hand side goes to ##(1-x)^{-2}## as ##h\to 0##. You either have to find a theorem that ensures that this is the case, or do an epsilon-delta proof.
 
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Related to Differential (derivative from first principles)

1. What is a differential (derivative from first principles)?

A differential (derivative from first principles) is a mathematical concept that represents the rate of change of a function at a particular point. It is calculated using the limit of the slope of a secant line as the two points get closer and closer together. This process is known as taking the derivative from first principles.

2. Why is the differential (derivative from first principles) important?

The differential (derivative from first principles) is important because it allows us to find the instantaneous rate of change of a function at a specific point. This has many real-world applications, such as in physics, engineering, and economics, where we often need to know how quickly a quantity is changing at a given moment.

3. How is the differential (derivative from first principles) calculated?

The differential (derivative from first principles) is calculated using the limit definition of the derivative, which involves finding the slope of a secant line between two points on the graph of the function and then taking the limit as those two points get closer and closer together. This process involves using algebraic manipulations and basic calculus rules.

4. What is the difference between a differential (derivative from first principles) and a derivative?

The differential (derivative from first principles) and the derivative are essentially the same thing. Both represent the instantaneous rate of change of a function at a particular point. However, the differential (derivative from first principles) is calculated using the limit definition of the derivative, while the derivative can also be calculated using other methods, such as the power rule, product rule, and chain rule.

5. Can the differential (derivative from first principles) be used to find the slope of a tangent line?

Yes, the differential (derivative from first principles) can be used to find the slope of a tangent line at a specific point on a curve. This is because the slope of a tangent line represents the instantaneous rate of change of the function at that point, which is exactly what the differential (derivative from first principles) measures.

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