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partial derivative of xe^y-y^2+e^x
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∂
∂
y
(
xe
y
−
y
2
+
e
x
)
=
e
y
x
−
2
y
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∂
∂
y
(
xe
y
−
y
2
+
e
x
)
=
e
y
x
−
2
y
steps
∂
∂
y
(
xe
y
−
y
2
+
e
x
)
Treat
x
as
a
constant
Apply
the
Sum
/
Difference
Rule
:
(
f
±
g
)
′
=
f
′
±
g
′
=
∂
∂
y
(
xe
y
)
−
∂
∂
y
(
y
2
)
+
∂
∂
y
(
e
x
)
show steps
∂
∂
y
(
xe
y
)
=
xe
y
show steps
∂
∂
y
(
y
2
)
=
2
y
show steps
∂
∂
y
(
e
x
)
=
0
=
xe
y
−
2
y
+
0
Simplify
=
e
y
x
−
2
y
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