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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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3 Answers 3

up vote 2 down vote accepted

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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You do not prove the existence of the derivative. –  mookid Mar 16 at 13:29
    
His answer is more clear! –  user135767 Mar 16 at 13:33
    
but it is not true. anyway. –  mookid Mar 16 at 16:06

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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That's not what I said. I said that Idon't understand the statement that he said. –  user135767 Mar 16 at 11:58
    
please re read my question –  user135767 Mar 16 at 11:59
    
I completed my answer. I you know the derivative for x=0, then you know it everywhere. –  mookid Mar 16 at 11:59
    
why? why if we know the derivative at x=0 then we know it everywhere? –  user135767 Mar 16 at 12:02
    
because of my argument. –  mookid Mar 16 at 12:50

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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