Why can the first derivative of the sigmoid function can be simplified as shown below
$begingroup$
Can you explain how the formula$$dfrac{mathrm{e}^{-x}}{left(mathrm{e}^{-x}+1right)^2}$$
which is the first derivative of the sigmoid function can be simplified into
$$dfrac{mathrm{e}^{x}}{left(mathrm{e}^{x}+1right)^2}$$
exponential-function
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
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add a comment |
$begingroup$
Can you explain how the formula$$dfrac{mathrm{e}^{-x}}{left(mathrm{e}^{-x}+1right)^2}$$
which is the first derivative of the sigmoid function can be simplified into
$$dfrac{mathrm{e}^{x}}{left(mathrm{e}^{x}+1right)^2}$$
exponential-function
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
$endgroup$
add a comment |
$begingroup$
Can you explain how the formula$$dfrac{mathrm{e}^{-x}}{left(mathrm{e}^{-x}+1right)^2}$$
which is the first derivative of the sigmoid function can be simplified into
$$dfrac{mathrm{e}^{x}}{left(mathrm{e}^{x}+1right)^2}$$
exponential-function
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
$endgroup$
Can you explain how the formula$$dfrac{mathrm{e}^{-x}}{left(mathrm{e}^{-x}+1right)^2}$$
which is the first derivative of the sigmoid function can be simplified into
$$dfrac{mathrm{e}^{x}}{left(mathrm{e}^{x}+1right)^2}$$
exponential-function
exponential-function
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
asked Dec 23 '18 at 20:18
Jean-Pierre SchnyderJean-Pierre Schnyder
1012
1012
add a comment |
add a comment |
4 Answers
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$begingroup$
Expand it out, and then multiply by $frac{e^{2x}}{e^{2x}}$:
$$
frac{e^{-x}}{(e^{-x}+1)^2}=frac{e^{2x}}{e^{2x}}cdotfrac{e^{-x}}{e^{-2x}+2e^{-x}+1}=frac{e^{x}}{1+2e^{x}+e^{2x}}=frac{e^{x}}{(e^{x}+1)^2}
$$
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
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$begingroup$
Since you mention this is the first derivative of $$sigma(x):=frac{1}{1+e^{-x}}=1-frac{e^{-x}}{1+e^{-x}}=1-frac{1}{1+e^x}=1-sigma(-x)$$(where the penultimate $=$ multiplies numerator and denominator by $e^x$),$$sigma'(x)=frac{d}{dx}(1-sigma(-x))=sigma'(-x)$$is an even function, which gives the desired result.
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
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$begingroup$
Multiply expression by $frac{ e^{2x} }{e^{2x} }$
The numerator clearly becomes $e^{2x-x} = e^x $ and the denominator
$$ (e^{-x}+1)^2 cdot (e^{x})^2 = ( (e^{-x}+1) e^x )^2 = (1 + e^x)^2 $$
answered Dec 23 '18 at 20:21
JamesJames
2,636425
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$begingroup$
$$frac{e^{-x}}{(1+e^{-x})^2}=frac{e^{-x}}{(1+e^{-x})^2}frac{e^{2x}}{e^{2x}}=frac{e^{-x+2x}}{((1+e^{-x})e^x)^2}=frac{e^{x}}{(e^x+1)^2}$$
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Expand it out, and then multiply by $frac{e^{2x}}{e^{2x}}$:
$$
frac{e^{-x}}{(e^{-x}+1)^2}=frac{e^{2x}}{e^{2x}}cdotfrac{e^{-x}}{e^{-2x}+2e^{-x}+1}=frac{e^{x}}{1+2e^{x}+e^{2x}}=frac{e^{x}}{(e^{x}+1)^2}
$$
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
$endgroup$
add a comment |
$begingroup$
Expand it out, and then multiply by $frac{e^{2x}}{e^{2x}}$:
$$
frac{e^{-x}}{(e^{-x}+1)^2}=frac{e^{2x}}{e^{2x}}cdotfrac{e^{-x}}{e^{-2x}+2e^{-x}+1}=frac{e^{x}}{1+2e^{x}+e^{2x}}=frac{e^{x}}{(e^{x}+1)^2}
$$
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
$endgroup$
add a comment |
$begingroup$
Expand it out, and then multiply by $frac{e^{2x}}{e^{2x}}$:
$$
frac{e^{-x}}{(e^{-x}+1)^2}=frac{e^{2x}}{e^{2x}}cdotfrac{e^{-x}}{e^{-2x}+2e^{-x}+1}=frac{e^{x}}{1+2e^{x}+e^{2x}}=frac{e^{x}}{(e^{x}+1)^2}
$$
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
$endgroup$
Expand it out, and then multiply by $frac{e^{2x}}{e^{2x}}$:
$$
frac{e^{-x}}{(e^{-x}+1)^2}=frac{e^{2x}}{e^{2x}}cdotfrac{e^{-x}}{e^{-2x}+2e^{-x}+1}=frac{e^{x}}{1+2e^{x}+e^{2x}}=frac{e^{x}}{(e^{x}+1)^2}
$$
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 20:21
RandomNumberGuyRandomNumberGuy
662
662
add a comment |
add a comment |
$begingroup$
Since you mention this is the first derivative of $$sigma(x):=frac{1}{1+e^{-x}}=1-frac{e^{-x}}{1+e^{-x}}=1-frac{1}{1+e^x}=1-sigma(-x)$$(where the penultimate $=$ multiplies numerator and denominator by $e^x$),$$sigma'(x)=frac{d}{dx}(1-sigma(-x))=sigma'(-x)$$is an even function, which gives the desired result.
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
$endgroup$
add a comment |
$begingroup$
Since you mention this is the first derivative of $$sigma(x):=frac{1}{1+e^{-x}}=1-frac{e^{-x}}{1+e^{-x}}=1-frac{1}{1+e^x}=1-sigma(-x)$$(where the penultimate $=$ multiplies numerator and denominator by $e^x$),$$sigma'(x)=frac{d}{dx}(1-sigma(-x))=sigma'(-x)$$is an even function, which gives the desired result.
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
$endgroup$
add a comment |
$begingroup$
Since you mention this is the first derivative of $$sigma(x):=frac{1}{1+e^{-x}}=1-frac{e^{-x}}{1+e^{-x}}=1-frac{1}{1+e^x}=1-sigma(-x)$$(where the penultimate $=$ multiplies numerator and denominator by $e^x$),$$sigma'(x)=frac{d}{dx}(1-sigma(-x))=sigma'(-x)$$is an even function, which gives the desired result.
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
$endgroup$
Since you mention this is the first derivative of $$sigma(x):=frac{1}{1+e^{-x}}=1-frac{e^{-x}}{1+e^{-x}}=1-frac{1}{1+e^x}=1-sigma(-x)$$(where the penultimate $=$ multiplies numerator and denominator by $e^x$),$$sigma'(x)=frac{d}{dx}(1-sigma(-x))=sigma'(-x)$$is an even function, which gives the desired result.
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
answered Dec 23 '18 at 20:25
J.G.J.G.
34k23252
34k23252
add a comment |
add a comment |
$begingroup$
Multiply expression by $frac{ e^{2x} }{e^{2x} }$
The numerator clearly becomes $e^{2x-x} = e^x $ and the denominator
$$ (e^{-x}+1)^2 cdot (e^{x})^2 = ( (e^{-x}+1) e^x )^2 = (1 + e^x)^2 $$
answered Dec 23 '18 at 20:21
JamesJames
2,636425
$endgroup$
add a comment |
$begingroup$
Multiply expression by $frac{ e^{2x} }{e^{2x} }$
The numerator clearly becomes $e^{2x-x} = e^x $ and the denominator
$$ (e^{-x}+1)^2 cdot (e^{x})^2 = ( (e^{-x}+1) e^x )^2 = (1 + e^x)^2 $$
answered Dec 23 '18 at 20:21
JamesJames
2,636425
$endgroup$
add a comment |
$begingroup$
Multiply expression by $frac{ e^{2x} }{e^{2x} }$
The numerator clearly becomes $e^{2x-x} = e^x $ and the denominator
$$ (e^{-x}+1)^2 cdot (e^{x})^2 = ( (e^{-x}+1) e^x )^2 = (1 + e^x)^2 $$
answered Dec 23 '18 at 20:21
JamesJames
2,636425
$endgroup$
Multiply expression by $frac{ e^{2x} }{e^{2x} }$
The numerator clearly becomes $e^{2x-x} = e^x $ and the denominator
$$ (e^{-x}+1)^2 cdot (e^{x})^2 = ( (e^{-x}+1) e^x )^2 = (1 + e^x)^2 $$
answered Dec 23 '18 at 20:21
JamesJames
2,636425
answered Dec 23 '18 at 20:21
JamesJames
2,636425
answered Dec 23 '18 at 20:21
JamesJames
2,636425
answered Dec 23 '18 at 20:21
JamesJames
2,636425
2,636425
add a comment |
add a comment |
$begingroup$
$$frac{e^{-x}}{(1+e^{-x})^2}=frac{e^{-x}}{(1+e^{-x})^2}frac{e^{2x}}{e^{2x}}=frac{e^{-x+2x}}{((1+e^{-x})e^x)^2}=frac{e^{x}}{(e^x+1)^2}$$
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
$endgroup$
add a comment |
$begingroup$
$$frac{e^{-x}}{(1+e^{-x})^2}=frac{e^{-x}}{(1+e^{-x})^2}frac{e^{2x}}{e^{2x}}=frac{e^{-x+2x}}{((1+e^{-x})e^x)^2}=frac{e^{x}}{(e^x+1)^2}$$
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
$endgroup$
add a comment |
$begingroup$
$$frac{e^{-x}}{(1+e^{-x})^2}=frac{e^{-x}}{(1+e^{-x})^2}frac{e^{2x}}{e^{2x}}=frac{e^{-x+2x}}{((1+e^{-x})e^x)^2}=frac{e^{x}}{(e^x+1)^2}$$
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
$endgroup$
$$frac{e^{-x}}{(1+e^{-x})^2}=frac{e^{-x}}{(1+e^{-x})^2}frac{e^{2x}}{e^{2x}}=frac{e^{-x+2x}}{((1+e^{-x})e^x)^2}=frac{e^{x}}{(e^x+1)^2}$$
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
answered Dec 23 '18 at 20:22
mrtaurhomrtaurho
6,19771641
6,19771641
add a comment |
add a comment |
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