Mixing
Write a Limit to calculate $f'(0)$
$begingroup$
Let $f(x) = frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx
((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx
((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx
(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)
(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)
(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)
that's as far as I have gotten. Any input would be great.
limits functions derivatives
asked Jan 22 at 19:23
user637291user637291
62
$endgroup$
|
show 2 more comments
$begingroup$
Let $f(x) = frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx
((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx
((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx
(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)
(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)
(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)
that's as far as I have gotten. Any input would be great.
limits functions derivatives
asked Jan 22 at 19:23
user637291user637291
62
$endgroup$
$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24
1
$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24
1
$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26
1
$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29
1
$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31
|
show 2 more comments
$begingroup$
Let $f(x) = frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx
((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx
((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx
(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)
(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)
(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)
that's as far as I have gotten. Any input would be great.
limits functions derivatives
asked Jan 22 at 19:23
user637291user637291
62
$endgroup$
Let $f(x) = frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx
((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx
((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx
(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)
(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)
(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)
that's as far as I have gotten. Any input would be great.
limits functions derivatives
limits functions derivatives
asked Jan 22 at 19:23
user637291user637291
62
asked Jan 22 at 19:23
user637291user637291
62
edited Jan 22 at 21:34
user637291
asked Jan 22 at 19:23
user637291user637291
62
asked Jan 22 at 19:23
user637291user637291
62
asked Jan 22 at 19:23
user637291user637291
62
62
$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24
1
$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24
1
$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26
1
$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29
1
$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31
|
show 2 more comments
$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24
1
$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24
1
$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26
1
$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29
1
$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31
$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24
$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24
1
1
$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24
$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24
1
1
$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26
$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26
1
1
$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29
$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29
1
1
$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31
$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
right) left( {x}^{2}+1 right) }}
$$
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
$endgroup$
$begingroup$
We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
$endgroup$
– The Jade Emperor
Jan 22 at 19:39
add a comment |
$begingroup$
$f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
$= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
right) left( {x}^{2}+1 right) }}
$$
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
$endgroup$
$begingroup$
We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
$endgroup$
– The Jade Emperor
Jan 22 at 19:39
add a comment |
$begingroup$
it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
right) left( {x}^{2}+1 right) }}
$$
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
$endgroup$
$begingroup$
We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
$endgroup$
– The Jade Emperor
Jan 22 at 19:39
add a comment |
$begingroup$
it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
right) left( {x}^{2}+1 right) }}
$$
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
$endgroup$
it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
right) left( {x}^{2}+1 right) }}
$$
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
answered Jan 22 at 19:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.5k42866
77.5k42866
$begingroup$
We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
$endgroup$
– The Jade Emperor
Jan 22 at 19:39
add a comment |
$begingroup$
We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
$endgroup$
– The Jade Emperor
Jan 22 at 19:39
$begingroup$
We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
$endgroup$
– The Jade Emperor
Jan 22 at 19:39
$begingroup$
We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
$endgroup$
– The Jade Emperor
Jan 22 at 19:39
add a comment |
$begingroup$
$f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
$= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
$endgroup$
add a comment |
$begingroup$
$f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
$= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
$endgroup$
add a comment |
$begingroup$
$f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
$= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
$endgroup$
$f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
$= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
edited Jan 22 at 19:39
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
answered Jan 22 at 19:30
Robert LewisRobert Lewis
47.9k23067
47.9k23067
add a comment |
add a comment |
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$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24
1
$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24
1
$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26
1
$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29
1
$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31