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Question 1
Mu
Derivatives
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Question
Evaluate
d
d
x
[
14
π
sin
(
x
)
]
∣
x
=
π
2
\left.\frac{d}{dx}\left[14\pi\sin(\sqrt{x})\right]\right|_{x=\pi^2}
d
x
d
[
14
π
sin
(
x
)
]
x
=
π
2
.
A.
−14
B.
−7
C.
7
D.
14
E.
NOTA
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