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Mathematics
Mathematics
Problem
For
x
≠
0
x\neq 0
x
=
0
, find
d
d
x
[
(
x
2
+
1
)
sin
x
x
]
\dfrac{d}{dx}\left[\dfrac{(x^{2}+1)\sin x}{x}\right]
d
x
d
[
x
(
x
2
+
1
)
sin
x
]
. a)
(
x
2
+
1
)
(
x
cos
x
−
sin
x
)
x
2
\dfrac{(x^{2}+1)(x\cos x-\sin x)}{x^{2}}
x
2
(
x
2
+
1
)
(
x
cos
x
−
sin
x
)
b)
x
(
x
2
+
1
)
cos
x
+
(
x
2
−
1
)
sin
x
x
2
\dfrac{x(x^{2}+1)\cos x+(x^{2}-1)\sin x}{x^{2}}
x
2
x
(
x
2
+
1
)
cos
x
+
(
x
2
−
1
)
sin
x
c)
x
(
x
2
+
1
)
cos
x
−
(
x
2
−
1
)
sin
x
x
2
\dfrac{x(x^{2}+1)\cos x-(x^{2}-1)\sin x}{x^{2}}
x
2
x
(
x
2
+
1
)
cos
x
−
(
x
2
−
1
)
sin
x
d)
x
(
x
2
+
1
)
cos
x
+
(
−
x
2
+
2
x
−
1
)
sin
x
x
2
\dfrac{x(x^{2}+1)\cos x+(-x^{2}+2x-1)\sin x}{x^{2}}
x
2
x
(
x
2
+
1
)
cos
x
+
(
−
x
2
+
2
x
−
1
)
sin
x
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