# Is my solution correct for the derivative of $\frac{\sin 3x}{\sqrt{3}}$?

Is my solution for the derivative of $\frac{\sin 3x}{\sqrt{3}}$ correct?

\begin{align} &\frac{d}{dx} \frac{\sin 3x}{\sqrt{3}} \\ & = \frac{0 \cdot \sin3x - \sqrt{3} \cdot 3\cos 3x}{(\sqrt{3})^2} \\ & = \frac{3 \cos 3x}{\sqrt{3}} \end{align}

I tried to verify through W|A but it says the answer is: $\sqrt{3} \cdot \cos 3x$. How can that be? Due to quotient rule, shouldn't there still be a $\sqrt{3}$ ?

EDIT:

Is it because $\frac{1}{\sqrt{3}}$ simplifies to $\frac{\sqrt{3}}{3}$ and the $3$'s cancel out?

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In case you misquoted W|A, recall $3/\sqrt 3 = \sqrt 3$. –  Lord_Farin Apr 9 '13 at 14:45
To begin with, you got the wrong sign in the quotient rule. Now it is absolutely unneceesary to use the quotient rule here. Just take the factor $\frac{1}{\sqrt{3}}$ out from the beginning. Differentiation is linear, never forget that. –  1015 Apr 9 '13 at 14:49

$$\frac{d}{dx} \frac{\sin 3x}{\sqrt{3}} = \frac d{dx}\left(\dfrac{1}{\sqrt 3} \cdot \sin 3x \right) = \frac1{\sqrt 3}\cdot 3 \cos 3x = \frac{3}{\sqrt 3} \cos 3x = \sqrt 3 \cos 3x \tag{1}$$

(And we can use the chain rule, only).

• Note that $\dfrac{1}{\sqrt 3}$ is a constant, so there's really no need to use the quotient rule. If you do use it, then the factors in your numerator should be reversed: $$d/dx\left(\dfrac{\sin 3x}{\sqrt{3}}\right) =\;\dfrac{(\sqrt 3)\cdot 3\cos 3x - 0\cdot \sin 3x}{(\sqrt 3)^2}.\;$$
• But aside from that, as you can see from $(1)$, the answer you arrive at equals the simplified answer given by W|A.

$(1)$ Note: $\; \dfrac 3{\sqrt 3} = \dfrac {\sqrt 3\cdot 3}{(\sqrt 3)^2} = \dfrac {3\cdot \sqrt 3}{3} = \sqrt 3,\;$ so yes, we can express/simplify and the $3$'s "cancel out".

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Hi, Amy. :-)... –  Babak S. Apr 9 '13 at 17:57
Hello, dear friend! I've been watching for you!! –  amWhy Apr 9 '13 at 17:57
I am like a pounded steak Amy. Exhausted... –  Babak S. Apr 9 '13 at 18:00
You deserve, then, to immerse yourself in a heavenly slumber! –  amWhy Apr 9 '13 at 18:02
Thanks so much for your kind words Amy. ;-) –  Babak S. Apr 9 '13 at 18:08

You for sure wrote it incorrectly .See this http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=d%2Fdx(sin3x%2Fsqrt%7B3%7D)

$$\dfrac1{\sqrt 3} \dfrac d{dx}(\sin3x)=\dfrac1{\sqrt 3}\times 3 \cos3x=\sqrt3 \cos3x$$

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A hint for formatting: putting a backslash before functions like $\sin$ gets them in the usual font. –  Ross Millikan Apr 9 '13 at 14:58