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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?

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because the derivative of a constant is 0, and thus that term cancels? –  Roger Apr 4 at 20:36
    
You can use the product rule if you look at the $3$ as the function $x\mapsto 3$. It is also legal to factor it out. –  Git Gud Apr 4 at 20:36
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how come you just wouldn't factor the constant out of all derivative questions?: you seem to be mistaken. We most definitely can. Could you show an example where you think we cannot?(not rhetoric, we need to see your doubt to clear it) –  Sabyasachi Apr 4 at 20:42
    
$$\frac{d}{dx}(3 f(x)) = 3 \frac{d}{dx} (f(x)) + f(x) \frac{d}{dx}(3) = 3 f'(x)+0 = 3 f'(x)$$ –  Henry Apr 4 at 20:47
    
@Henry Wow, this stuff is like magic! –  chopper draw lion4 Apr 4 at 20:54

2 Answers 2

up vote 5 down vote accepted

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 $a,b$.

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Eureka! I have asked many professors at my University this question and all of them gave conflicting answers. This is the best one I have heard. Everything makes sense now and I can rest peacefully. Thank you @mixedmath –  chopper draw lion4 Apr 4 at 21:14

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

$$(a\cdot f(x)+b\cdot g(x))' = a\cdot f'(x) + b\cdot g'(x)$$

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