Ignoring constants when finding derivatives of trig functions Find the derivative of $3sin^2(6x)$. I know I solve this by using chain rule, my question is how come product rule is not used instead on $3$ and $sin^2(6x)$? According to wolfram alpha, the constant is factored out at the beginning of the problem, if that is the case: how come you just wouldn't factor the constant out of all derivative questions?
 A: If you'd like to use the product rule, then use the product rule. Let's do it now.
The $(fg)' = f'g + fg'$, so 
$$\frac{d}{dx} 3\sin^2(6x) = \left(\frac{d}{dx} 3\right) \sin^2(6x) + 3\left(2\sin(6x)\cos(6x)6\right) = 0 + 3\left(12\sin(6x)\cos(6x)\right).$$
You'll notice that it's just as if the $3$ factored out. While I'd recommend recognizing that you can pull constants out, it is entirely possible to use the product rule every time.
You also ask: How come you wouldn't always pull the constant out? My answer - I always would pull the constant out. Let's even prove that this happens:
A derivative of a function $f$ is
$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h},$$
so that
$$ \frac{d}{dx} cf(x) = \lim_{h \to 0} \frac{cf(x+h) - cf(x)}{h} = c\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = cf'(x).$$
More generally, differentiation is a linear operator, meaning that $(af(x) + bg(x))' = af'(x) + bg'(x)$ for any functions $f,g$ and constants $a,b$.
A: As @GitGud points out, you can use the product rule. But the result is the same every time. If $c$ is a constant and $f$ is a function, then you always find that, using the product rule (and the fact that $c$ has a derivative of $0$)
$$\boxed{(c\cdot f(x))'} = c'\cdot f(x) + c\cdot f'(x) = 0\cdot f(x) + c\cdot f'(x)=\boxed{c\cdot f'(x)}$$
In other words, as you also asked, you can indeed pull any constant factor out front and take the derivative of what remains.
Examples: 
$$(7x^2)'=7(x^2)' = 7(2x)=14x$$
$$\left(\frac23\sin5x\right)'=\frac23(\sin5x)' = \frac23(\cos 5x\cdot 5)=\frac{10}{3}\cos5x$$
$$(3\sin^2 6x)'= 3(\sin^2 6x)' = 3(2\sin 6x\cdot\cos6x\cdot6)=36\sin 6x\cos6x
$$
It is even more generally true that differentiation is a linear operator, in the sense that for constants $a$ and $b$ it is true that
$$(a\cdot f(x)+b\cdot g(x))' = a\cdot f'(x) + b\cdot g'(x)$$
