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Could someone provide a basic proof that $\frac{d}{dx}(\sin(nx))=n\cos(nx)$? I'm using $n$ to be broad, and so this can be searched easier, though if it's easier to provide an example, then replace $n$ with $3$.
My teacher has just taught us the Chain Rule, and he showed us an example like this:

$$f(x)=(\sin(3x))^3$$ $$f'(x)=3(\sin(3x))^2(\cos(3x))(3)$$ $$f'(x)=9\sin^2(3x)\cos(3x)$$

I understand that the derivative of $\sin(x)$ is $\cos(x)$ and I understand that the derivative of $3x$ is $3$, but I just don't see how this makes sense, because it feels like you're taking the derivative of $\sin(3x)$ which as far as I can tell should be $\cos(3x)$, so where did the 3 come from? I was told that the 3 was taken as the derivative of the 3x inside the $\cos$... but why?
Isn't the $\cos(3x)$ already differentiated?

My teacher made a few different attempts to explain this to me, and I just don't get it, this is why I've asked for a proof. Now, I ask for a basic proof, because I don't want one of these super textbook like answers. Feel free to cut corners like not always putting $f(x)$ in front of each line or something like that, I just want to wrap my head around this concept!

Oh and, I've already graphed $\sin(3x)$ and $3\cos(3x)$, which so far is the only way I was able to actually see that, that is indeed the derivative.

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    $\begingroup$ So, do you feel that the derivative of $\sin(0x)$ should be $\cos(0x)$? $\endgroup$ – Somos Oct 24 '18 at 4:17
  • $\begingroup$ Kind of subtle but $f'(3x)=\lim \frac {f (3x+3h)-f (3x)}{3h}\ne \lim \frac {f (3x+3h)-f (3x)}{h}=3\lim\frac {f (3x+3h)-f (3x)}{3h}=3\lim\frac {f(3x+h)-f (3x)}h=[f (3x)]'$. (Replace $3h$ with $h $ to see the second to last step). So $[f (3x)]'=3f'(3x) $ This has nothing to do with trig and everything to do with the chain rule. $\endgroup$ – fleablood Oct 24 '18 at 4:35
  • $\begingroup$ This is not a proof and these are not fractions so you should not have expected them to cancel like this... but you have $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}$. Similarly $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dv}\dfrac{dv}{dx}$. If you were to naively treat these as fractions, the chain rule would appear to be straightforward algebraic cancellations, and there is reason for that. $\endgroup$ – JMoravitz Oct 24 '18 at 4:45
  • $\begingroup$ $\sin (nx)$ will change $\cos (nx)$ times as fast as $nx$ changes. And $nx$ will change $n$ times as fast as $x$ changes. So $\sin(nx)$ will change $\cos(nx) \times n$ times as fast as $x$ changes. $\endgroup$ – fleablood Oct 24 '18 at 5:16
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This is a simple matter of not understanding the chain rule, that’s all (it’s hard!).

When you say that the derivative of $\sin({3x})$ is $\cos(3x)$, that is actually true! - but only if you’re differentiating with respect to $3x$.

Observe:

$\frac{d}{dy}\sin(y) = \cos(y)$

Let $y=3x$. Then

$\frac{d}{d(3x)}\sin(3x) = \cos(3x)$

If you differentiate with respect to the thing you’re taking the $\sin$ of, you get $\cos$ of that thing; that is, when the variable in denominator of $\frac{d}{dy}$ (which is $y$ right here!) exactly matches the variable that the function $\sin(y)$ is of (also $y$ right here!), then the derivative is just as you calculated: $\cos(y)$.

But you’re NOT asked to differentiate with respect to $3x$; you’re asked to differentiate with respect to $x$! Since $x$ is NOT the thing we're taking $\sin$ of (it's $3x$!), the Chain Rule tells us that after we differentiate with respect to $3x$, we must multiply by the derivative of $3x$ with respect to $x$; that is, $3$.

Again, observe:

$\frac{d}{d\color{blue}{x}} \sin(\color{red}{3x}) = \frac{d}{d(\color{red}{3x})}\sin(\color{red}{3x}) \cdot \frac{d}{d\color{blue}{x}} (3\color{blue}{x})$

Since $\frac{d}{d(\color{red}{3x})}\sin(\color{red}{3x}) = \cos(\color{red}{3x})$, and $\frac{d}{d\color{blue}{x}}(3\color{blue}{x}) = 3$, then

$\frac{d}{dx} \sin(3x) = (\cos(3x)) \cdot 3$

Edit:

When I say "with respect to", technically what I mean is "how does this function change as this one variable changes". But you don't really need to fully wrap your head around that to be able to use the Chain Rule. Let's just say it means "the 'something' in $\frac{d}{d(something)}$".

For example, let's say we wanted to find the derivative of the function $ax^2$ with respect to $x$. That is, we want to calculate $\frac{d}{dx} ax^2$. That means that we will treat $x$ as the variable and $a$ as some constant (as if it were the number $3$). By the Power Rule, this would differentiate to $2ax$.

However, we could also make an assumption that $a$ is the variable in this function, and we want to find the derivative with respect to $a$. That is, we want to calculate $\frac{d}{da} ax^2$. (Notice $a$ is now the 'something' in $\frac{d}{d(something)}$). In this instance, we will treat $a$ as the variable and $x^2$ as some constant (as if it were the number $3$). Now this thing is just a constant times a variable, so it differentiates to the constant $x^2$.

Finally, let's consider what would happen if we treated the whole of $x^2$ as the variable (rather than just $x$), and we want to find the derivative of $ax^2$ with respect to $x^2$; i.e. to find $\frac{d}{d(x^2)} ax^2$. This means that we'll once again treat $a$ as constant, but now $x^2$ will be treated as the entire variable. To calculate this, we have a constant times a variable, which merely differentiates to a constant - meaning $\frac{d}{d(x^2)} ax^2 = a$.

Does this help?

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  • $\begingroup$ Very good and to the point answer. The only thing I thought you could do better is write $\frac {d(sin(3x))}{d(3x)}$, that way OP could see how the numerator and the denominator cancel giving us what we desire, but it is just a matter of preference. Regardless, a very good answer. $\endgroup$ – Bertrand Wittgenstein's Ghost Oct 24 '18 at 7:53
  • $\begingroup$ I don't really know what you mean to differentiate 'with respect' to something. In my class, we just started the unit on derivatives. I understand that with the chain rule I also have to get the derivative of what's inside the trig function I just didn't really see why. I think this is a case of, I was interpreting trigonometric derivatives exactly as I saw them. An example of what I mean is: I'm told that the derivative of $sin(x)$ is $cos(x)$, which even as you stated, is true. But maybe a better way to teach this to people is to use the form of derivative of $sin(nx)$ = $ncos(nx)$. $\endgroup$ – Magicrafter13 Gaming Oct 24 '18 at 17:10
  • $\begingroup$ (Continue) But I suppose that is just me having a preference, and isn't an actual flaw in how this is taught, so there's really nothing I can do about it. $\endgroup$ – Magicrafter13 Gaming Oct 24 '18 at 17:11
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    $\begingroup$ Yes, but I'm guessing your teacher wants you to understand $why$ the derivative of $\sin(nx) = n\cos(nx)$. And this is a perfect basic example of the Chain Rule. Please see the addition to my solution. $\endgroup$ – bloomers Oct 24 '18 at 18:14
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Are you familiar with Taylor series?

$\sin(nx)=(nx)-\frac{(nx)^3}{3!}+\frac{(nx)^5}{5!}-\frac{(nx)^7}{7!}+\cdots$

$\cos(nx)=1-\frac{(nx)^2}{2!}+\frac{(nx)^4}{4!}-\frac{(nx)^6}{6!}+\cdots$

$\sin'(nx)=n-n\frac{(nx)^2}{2!}+n\frac{(nx)^4}{4!}-n\frac{(nx)^6}{6!}+\cdots=n \cos(nx)$

A more intuitive explanation: Think the $3$ in $\sin(3x)$ as a factor that compresses the function horizontally by a factor of $3$. Therefore, where used to have a "steepness" of 1, due to the compression, must now have a "steepness" of 3.

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COnsider:

If we let $g(x) = f(3x)$

$g'(x) = \lim\limits_{h\to 0} \frac {f(3(x+h)) - f(3x)}{h}= $

$\lim\limits_{h\to 0} \frac {f(3x + 3h) - f(3x)}{h} = \lim\limits_{j=3h \to 0} \frac {f(3x + j) - f(3x)}{\frac 13j}=$

$3\cdot\lim\limits_{j \to 0} \frac {f(3x + j) - f(3x)}{ j}=3\cdot f'(3x)$

...

The subtle thing is that in taking the derivative of $\sin (nx)$ we are not taking the rate of change in $\sin (nx)$ with regard to the rate of change in $nx$. (That would be $\sin'(nx)= \cos(nx)$.) But instead we are taking the rate of change in $\sin (nx)$ with regard to the rate of change in $x$. And that change is $n$ times faster than in regard to rate of change in regard to $nx$. (because $\sin (nx)$ changes $\cos(nx)$ times as fast as $nx$ changes, but $nx$ changes $n$ times as fast as $x$ changes-- so $sin(nx)$ changes $\cos(nx) \times n$ times as fast as $x$ changes.)

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Your question seems to be mainly concerned about why the derivative of $\sin nx$ is $n \cos nx$ (as opposed to why $\sin nx = n \cos nx$. It might be easier to see that $\frac{d}{dx}(\sin(f(x)) = \cos(f(x)) \cdot f'(x)$, using Chain Rule. Now, let $f(x) = nx$, which gives us $\frac{d}{dx}\sin (nx) = \cos (nx) \cdot (\frac{d}{dx}(nx)) = \cos(nx) \cdot n = n\cos nx$.

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Actualy, you can compute the diffferential of $sin(nx)$ by definition. $sin'(nx)=\lim_{\triangle x\to 0} \frac{sin(nx+n\triangle x)-sin(nx)}{\triangle x}=\lim_{\triangle x\to 0} \frac{sin(nx)cos(n\triangle x)+sin(n\triangle x)cos(nx)-sin(nx)}{\triangle x}$

since $lim_{\triangle x\to0}cos(n\triangle x)=1,\qquad lim_{\triangle x\to0}sin(n\triangle x)=nx$

we finally have $sin'(nx)=ncos(nx)$.

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$$\sin'\big(nx\big)=\lim_{h\to 0}\frac{\sin\big(n(x+h)\big)-\sin(nx)}{h}$$ $$\lim_{h\to 0}\frac{\sin(nx)\cos(nh)+\cos(nx)\sin(nh)-\sin(nx)}{h}$$ $$\lim_{h\to 0}\frac{\sin(nx)\cos(nh)-\sin(nx)}{h}+\lim_{h\to0}\frac{\cos(nx)\sin(nh)}{h}$$ $$\lim_{h\to 0}\frac{n\sin(nx)\big(\cos(nh)-1\big)}{nh}+\lim_{h\to0}\frac{n\cos(nx)\sin(nh)}{nh}$$

Now: $$\lim_{h\to 0}\frac{n\sin(nx)\big(\cos(nh)-1\big)}{nh}=0$$ because $\frac{\cos(nh)-1}{nh}\to 0$ as $h\to 0$.

On the other hand

$$\lim_{h\to0}\frac{n\cos(nx)\sin(nh)}{nh}=n\cos(nx)$$

because $\frac{\sin{nh}}{nh}\to 1$ as $h\to0$.

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