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In video 3 of the video lectures by MIT on Single Variable Calculus presented by David Jerison, the latter says:

Remarks:
$\dfrac{d}{dx}\cos x\left|\right._{x=0}=\lim\limits_{\Delta x\to0}\dfrac{\cos\Delta x-1}{\Delta x}=0$
$\dfrac{d}{dx}\sin x\left|\right._{x=0}=\lim\limits_{\Delta x\to0}\dfrac{\sin\Delta x}{\Delta x}=1$

Okay I understand this, but then he says:

Derivatives of sine and cosine at %x=0% give all values of $\dfrac{d}{dx}\sin x$ and $\dfrac{d}{dx}\cos x$.

What?? What does he mean?

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mookid's answer is fine. Or try this. Suppose you want to compute the derivative of $\cos$ at a point $a$. Use the identity $$ \cos(a+x)=\cos(a)\cos(x)-\sin(a)\sin(x) . $$ Differentiate that with respect to $x$, $$ \cos'(a+x)=\cos(a)\cos'(x)-\sin(a)\sin'(x) , $$ then plug in $x=0$ $$ \cos'(a)=\cos(a)\cos'(0)-\sin(a)\sin'(0) . $$ Now if you know $\sin'(0)=1$ and $\cos'(0)=0$, you get $$ \cos'(a) = -\sin(a) . $$

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  • $\begingroup$ You do not prove the existence of the derivative. $\endgroup$ – mookid Mar 16 '14 at 13:29
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    $\begingroup$ His answer is more clear! $\endgroup$ – user135767 Mar 16 '14 at 13:33
  • $\begingroup$ but it is not true. anyway. $\endgroup$ – mookid Mar 16 '14 at 16:06
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He makes use of $$ \cos(x+h) - \cos(x) = -2\sin (x+h/2)\sin(h/2)\\ \sin(x+h) - \sin(x) = 2\cos (x+h/2)\sin(h/2)\\ $$

then $$ \lim_{h\to 0} \frac 1h [\cos(x+h) - \cos(x)] = -\sin (x) \lim_{h\to 0} \frac 2h\sin(h/2) = -\sin (x)\frac d{dx}\sin(x)|_{x=0} \\ \lim_{h\to 0} \frac 1h [\sin(x+h) - \sin(x)] = \cos (x)\lim_{h\to 0} \frac 2h\sin(h/2) =\cos(x) \frac d{dx}\sin(x)|_{x=0} $$ and you know that

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  • $\begingroup$ That's not what I said. I said that Idon't understand the statement that he said. $\endgroup$ – user135767 Mar 16 '14 at 11:58
  • $\begingroup$ please re read my question $\endgroup$ – user135767 Mar 16 '14 at 11:59
  • $\begingroup$ I completed my answer. I you know the derivative for x=0, then you know it everywhere. $\endgroup$ – mookid Mar 16 '14 at 11:59
  • $\begingroup$ why? why if we know the derivative at x=0 then we know it everywhere? $\endgroup$ – user135767 Mar 16 '14 at 12:02
  • $\begingroup$ because of my argument. $\endgroup$ – mookid Mar 16 '14 at 12:50
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I didn't watch the video, so this might not be on point. But just compute the derivative of $\sin x$ at the point $x = a$:

$$\left.{d\over dx}\sin{x}\right|_{x=a} = \ \lim_{h\rightarrow 0} \frac{\sin{\!(a + h)}-\sin{a}}{h}\,.$$

Using the identity $\sin{\!(a + h)} = \sin{a}\cos{h} + \cos{a}\sin{h}$,

$$\left.{d\over dx}\sin{x}\right|_{x=a} = \ \lim_{h\rightarrow 0} \frac{\sin{a}\cos{h} + \cos{a}\sin{h}-\sin{a}}{h}$$

$$ = \cos{a}\left(\lim_{h\rightarrow 0} \frac{\sin{h}}{h}\right) + \sin{a}\left(\lim_{h\rightarrow 0} \frac{\cos{h} - 1}{h}\right)\,.$$

Perhaps now you can see that if we know the two principal limits at $0$, we obtain $(\sin{x})^\prime$ at any value of $x$.

Was that your question? You might try this with $(\cos{x})^\prime$ as an exercise.

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Jason Zimba's answer actually is the one that explains the most about the particular subtlety of the statement in the video.

In the video before the moment in question, the lecturer derives formulae of $ sin'(x) $ and $cos'(x)$ based upon the (yet unproven) facts which he denotes as $A$ and $B$:

\begin{align} \lim_{\Delta x \to 0} {\frac{\cos \Delta x - 1}{\Delta x}} = 0 \tag{A} \\ \lim_{\Delta x \to 0} {\frac{\sin \Delta x}{\Delta x}} = 1 \tag{B} \end{align}

Then he says that we surely need to prove $A$ and $B$ in order for these derivations to be valid.

In other words, he says that $$ A, B \implies \text{proofs of the derivations of sin' and cos'} $$ (i.e. if we prove A, B then the proofs are sound)

Then, in the remark in question, the lecturer is considering the derivatives of the sine and cosine at $x = 0$ using the definition of a derivative and derives that: \begin{align} \dfrac{d}{dx} \cos x |_{x=0} &= \lim_{\Delta x \to 0} {\frac{\cos \Delta x - 1}{\Delta x}} \\ \dfrac{d}{dx} \sin x |_{x=0} &= \lim_{\Delta x \to 0} {\frac{\sin \Delta x}{\Delta x}} \end{align}

From which we can see that the derivatives at $0$ equal to the left hand sides of the equations $A$ and $B$ that we need to prove.

Hence, if we'd be able to prove that \begin{align} \dfrac{d}{dx} \cos x |_{x=0} &= 0 \\ \dfrac{d}{dx} \sin x |_{x=0} &= 1, \end{align} by transitivity the statements $A$ and $B$ will be proven.

I think that in the lecture the last idea was formulated a bit unclear, particularly the lecturer said that "the derivatives at $0$ give us all the values of the derivatives" while effectively he meant that the derivatives at $0$ give us the values of l.h.s. terms in A and B which in turn give us the formulae for $sin'(x)$ and $cos'(x)$

Basically, David Jarison has just used the wording "all the values of the derivatives" in place of "analytical formulae for $sin(x)$ and $cos(x)$".

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