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partial derivative of 1/(1+e^{-x})
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∂
∂
x
(
1
1
+
e
−
x
)
=
e
−
x
(
1
+
e
−
x
)
2
Show Steps
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∂
∂
x
(
1
1
+
e
−
x
)
=
e
−
x
(
1
+
e
−
x
)
2
steps
∂
∂
x
(
1
1
+
e
−
x
)
Apply
exponent
rule
:
1
a
=
a
−
1
=
∂
∂
x
(
(
1
+
e
−
x
)
−
1
)
show steps
Apply
the
chain
rule
:
−
1
(
1
+
e
−
x
)
2
∂
∂
x
(
1
+
e
−
x
)
=
−
1
(
1
+
e
−
x
)
2
∂
∂
x
(
1
+
e
−
x
)
show steps
∂
∂
x
(
1
+
e
−
x
)
=
−
e
−
x
=
−
1
(
1
+
e
−
x
)
2
(
−
e
−
x
)
show steps
Simplify
−
1
(
1
+
e
−
x
)
2
(
−
e
−
x
)
:
e
−
x
(
1
+
e
−
x
)
2
=
e
−
x
(
1
+
e
−
x
)
2
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