\(\frac{1}{3} \sin 3x\)
\(-\frac{1}{3} \sin 3x\)
3 sin 3x
-3 sin 3x
Correct answer is D
y = cos 3x
Let u = 3x so that y = cos u
Now, \(\frac{\delta y}{\delta x} = 3\),
\(\frac{\delta y}{\delta x} = -sin u\)
By the chain rule,
\(\frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x}\)
\(\frac{\delta y}{\delta x} = (-\sin u) (3)\)
\(\frac{\delta y}{\delta x} = -3 \sin u\)
\(\frac{\delta y}{\delta x} = -3 \sin 3x\)
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