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$g=\sin$, $g'=\cos$

If $g(x)=\sin(x)$, then $g'(x)=\cos(x)$.

Then $g(y)=\sin(y)$ and $g'(y)=\cos(y)$.

Let $y=2x$.

Then $g(y)=\sin(2x)$ and $g'(y)=\cos(2x)$, but if $h(x)=\sin(2x)$, then $h'(x)=2\cos(2x)$,
so $h(x)=g(y)$, but $h'(x)$ is not equal to $g'(y)$ if, say, $x = \pi$.

This doesn't make sense to me, isn't to say $h(x)=g(y)$ to say that $h(x)$ and $g(y)$ are the same? and so wouldn't it follow that $h'(x)=g'(y)$?

Is my problem that I am looking at individual values, $h(x)$ and $g(y)$, instead of the functions $h$ and $g$ themselves?

Edit: Once I thought to think of derivative as slope and realized that the slope of the line tangent to the graph of $h(x)$ at $x=a$ is not necessarily the same as the slope of the line tangent to the graph of $g(x)$ at $x=a$, I understood (satisfactorily) why $g'(x)$ isn't the same as $h'(x)$.

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You are using primes to denote two different things, which we can loosely describe as "derivatives with respect to x" and "derivatives with respect to y" –  Mariano Suárez-Alvarez Oct 14 '12 at 0:01
    
@Mariano: The primes mean the same thing in both cases: "The derivative of the univariate function that is decorated with the prime". –  Hurkyl Oct 14 '12 at 3:37
    
they cannot mean that, for otherwise, for example, the equalities $g(y)=\sin(2x)$ and $g'(y)=\cos(2x)$ are simply incompatible. –  Mariano Suárez-Alvarez Oct 14 '12 at 3:49
    
What's the problem? $g' = \cos$, so $g'(y) = \cos y = \cos(2x)$ –  Hurkyl Oct 14 '12 at 3:50
    
Hurkyl is correct about the meaning of "prime". However, there is implicit equivocation involved in the OP's conclusion. (See my answer.) –  Cameron Buie Oct 14 '12 at 4:42
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6 Answers 6

Try using chain rule: $$ (\sin(\varphi(t)))'_{t}=\varphi'(t) \cos(\varphi(t)) $$ Setting $\varphi(t)=2t$ you get $(\sin(\varphi(t)))'_{t}=2 \cos2t$

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If you compare the graph of $y=\sin(x)$ to that of $y=\sin(2x)$, the latter is compressed toward the $y$ axis by a factor of $1/2$ compared to the former. This contraction definitely will double slopes, at least if such slopes aren't $0$.

But $\cos(2x)$ never reaches above $1$ or below $-1$, so some points on $y=\sin(2x)$ will have slopes unreachable by $\cos(2x)$.

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Please \cos x gives you $\cos x$, whereas cos x gives you $cos x$. Both spacing and italicization are affected. –  Michael Hardy Oct 14 '12 at 4:03
    
A minor point: if the original slope is $0$, then so is the doubled slope, so it still works even then. –  Cameron Buie Oct 14 '12 at 4:39
    
Cameron: Yes, twice zero is zero. But it's not zero!. Because zero! means $0!$ which is 1. –  coffeemath Oct 14 '12 at 5:24
    
Michael: Thanks for the edit!. I do know some LaTex but some of the fine points still elude me. I guess writing $cosx$ gives something not as good as $\cos x$. Hmm: just now I included the flanking dollar signs and even the comment was the LaTex version. Probably should have written \frac{1}{2} also. –  coffeemath Oct 14 '12 at 5:26
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Given a function $f$ in one variable--say $t$, for example--we use $f'(t)$ to denote $\frac{d}{dt}\bigl[f(t)\bigr],$ so we're taking the derivative with respect to the given variable.

Now, let me translate the problematic point into the other notation, so that I can clarify the problem:

"This doesn't make sense to me, isn't to say $h(x)=g(y)$ to say that $h(x)$ and $g(y)$ are the same? and so wouldn't it follow that $\frac{d}{dx}\bigl[h(x)\bigr]=\frac{d}{dy}\bigl[g(y)\bigr]$?"

The answer to the first question is yes. To the second, however, the answer is no. Now, if we took the derivative on both sides with respect to the same variable, then the latter answer would also be yes. However, to do this correctly, we'll probably need to use the chain rule.

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Taking $\,f(x):=\sin 2x\,$ , we get:

$$\sin 2x-\sin 2x_0=2(\sin x\cos x-\sin x_0\cos x_0)=$$

$$=2(\sin x\cos x-\sin x\cos x_0+\sin x\cos x_0-\sin x_0\cos x_0)$$

Thus, by the definition of derivative, using the above, using the formulae for the double angle ($\,\sin 2x=2\sin x\cos x\,\,,\,\,\cos 2x=\cos^2x-\sin^2x\,$) and taking into account

that $\,(\sin x)'=\cos x\,\,,\,\,(\cos x)'=-\sin x\,$ , we get:

$$f'(x_0):=\lim_{x\to\ x_0}\frac{\sin 2x-\sin 2x_0}{x-x_0}=2\lim_{x\to x_0}\,\left[\sin x\frac{\cos x-\cos x_0}{x-x_0}+\cos x_0\frac{\sin x-\sin x_0}{x-x_0}\right]=$$

$$=2\left[\sin x_0(-\sin x_0)+\cos x_0(\cos x_0)\right]=2(\cos^2x_0-\sin^2x_0)=2\cos 2x_0$$

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Just to add another perspective: It is true that $\sin'(x) = \cos(x)$. It therefore follows that $\sin'(2x) = \cos(2x)$. However, the derivative of $\sin(2x)$ is not $\sin'(2x)$. The function $\sin(2x)$ is the composite of $x\mapsto 2x$ and $x\mapsto\sin(x)$, and the chain rule tells you how both parts contribute to the derivative of the total: $$\frac{d}{dx}\sin(2x) = \left(\frac{d}{dx}(2x)\right)\sin'(2x) = 2\sin'(2x) = 2\cos(2x).$$

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It is true that if if $h = g$, then $h' = g'$.

Equality of functions is given pointwise: $h=g$ is true if and only if $h(a) = g(a)$ for every real number $a$.

However, $h=g$ is not true: $h(a) = \sin(2a)$ but $g(a) = \sin a$, and these are usually different.

Now, $h(x) = g(y)$ is indeed true: both sides express the same functional dependence on $x$, or equivalently of $y$. Correspondingly, their differentials are the same:

$$ d(h(x)) = d(g(y)) $$ $$ h'(x) \,dx = g'(y) \,dy $$ $$ 2 \cos 2x\,dx = \cos y \,dy $$

Since $y = 2x$, we also have $dy = 2 dx$, and so everything works out.

Of course, if you're not comfortable with differentials, all of the above remains the same if you pick a specific variable to differentiate both sides with respect to. i.e. if $x$ and $y$ are both functions of $t$, then we have

$$ \frac{d}{dt} h(x) = \frac{d}{dt} g(y) $$ $$ \frac{dy}{dt} = 2 \frac{dx}{dt} $$

and we can do a similar calculation.

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