Derivative of $3 \cdot sin(2x)$

Question

Is the derivative of $3 \cdot sin(2x)$

$3 \cdot [cos(2x) \cdot 2]$ or $3\cdot [cos(2x)]\cdot 2$?

I'm unsure about this technicality. The first one seems more reasonable to me.

Both derivatives you gave are equal to $6\cos(2x)$. – E W H Lee Sep 28 '14 at 17:34 — E W H Lee, Sep 28 '14 at 17:34

score 3 · Accepted Answer · answered Sep 28 '14 at 17:35

3

Remember that $$a \cdot (b \cdot c) = (a \cdot b) \cdot c = a \cdot b \cdot c = a \cdot c \cdot b$$ and so on. Laws of multiplication.

Therefore $3[\cos(2x) \cdot 2] = 3 \cdot \cos(2x) \cdot 2 = 6 \cdot \cos (2x)$

answered Sep 28 '14 at 17:35

Sheheryar Zaidi

score 0 · Answer 2 · answered Sep 28 '14 at 17:38

0

Both of then are correct

$ 6 \cdot \cos (2x) = 3\cdot [\cos(2x) \cdot 2] = 3 \cdot \cos(2x) \cdot 2 $

answered Sep 28 '14 at 17:38

2 Answers2