Differential (derivative from first principles)

  • #1
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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Answers and Replies

  • #2
ehild
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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
Dick
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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
Fredrik
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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.

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
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 Im 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
Fredrik
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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 Im 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
Fredrik
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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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