I don't know how to take out Δx when i do the question

If $y=sin x$, show that

$
\frac{dy}{dx}= cos x
$

(by the first principles)

2. Originally Posted by ling_c_0202
I don't know how to take out Δx when i do the question

If $y=sin x$, show that

$
\frac{dy}{dx}= cos x
$

(by the first principles)
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{sin(x + \Delta x) - sin(x)}{\Delta x}$

Now, $sin(A + B) = sin(A)cos(B) + sin(B)cos(A)$:

$= \lim_{\Delta x \to 0} \frac{sin(x)cos( \Delta x) + sin( \Delta x)cos(x) - sin(x)}{\Delta x}$

$= \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} + \frac{sin( \Delta x)cos(x)}{ \Delta x} - \frac{sin(x)}{\Delta x} \right )$

$= \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} \right )$ + $\lim_{\Delta x \to 0} \left ( \frac{sin( \Delta x)cos(x)}{ \Delta x} \right )$ - $\lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )$

Now, $\lim_{\Delta x \to 0} \frac{sin( \Delta x)}{ \Delta x} = 1$ and $\lim_{\Delta x \to 0} cos( \Delta x) = 1$ so:

$\frac{dy}{dx} = \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{ \Delta x} \right ) + cos(x) - \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )$

$= cos(x)$.

-Dan

3. Originally Posted by topsquark
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{sin(x + \Delta x) - sin(x)}{\Delta x}$

Now, $sin(A + B) = sin(A)cos(B) + sin(B)cos(A)$:

$= \lim_{\Delta x \to 0} \frac{sin(x)cos( \Delta x) + sin( \Delta x)cos(x) - sin(x)}{\Delta x}$

$= \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} + \frac{sin( \Delta x)cos(x)}{ \Delta x} - \frac{sin(x)}{\Delta x} \right )$

$= \lim_{\Delta x \to 0} \left ( \frac{sin(x)cos( \Delta x)}{ \Delta x} \right )$ + $\lim_{\Delta x \to 0} \left ( \frac{sin( \Delta x)cos(x)}{ \Delta x} \right )$ - $\lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )$
You can't do it this way as you are going to have to subtract infnities
to get the finite limit - maybe OK in Physics but not here

Now, $\lim_{\Delta x \to 0} \frac{sin( \Delta x)}{ \Delta x} = 1$ and $\lim_{\Delta x \to 0} cos( \Delta x) = 1$ so:

$\frac{dy}{dx} = \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{ \Delta x} \right ) + cos(x) - \lim_{\Delta x \to 0} \left ( \frac{sin(x)}{\Delta x} \right )$

$= cos(x)$.

-Dan
RonL

4. Originally Posted by CaptainBlack
You can't do it this way as you are going to have to subtract infnities
to get the finite limit - maybe OK in Physics but not here

RonL
Doh! Sorry!

-Dan

5. Let die Meister try.
---
You need to find, for the point $x$.
$\lim_{\Delta x\to 0}\frac{\sin(x+\Delta x)-\sin x}{\Delta x}$
$\lim_{\Delta x\to 0}\frac{\sin x\cos \Delta x+\cos x \sin \Delta x-\sin x}{\Delta x}$
$\lim_{\Delta x\to 0}\left[\sin x\left( \frac{\cos \Delta x -1}{\Delta x}\right)+\cos x\left( \frac{\sin \Delta x}{\Delta x} \right) \right]$
Recognize the famous limits,
$\lim_{\Delta x\to 0}\frac{\sin \Delta x}{\Delta x}=1$
$\lim_{\Delta x\to 0}\frac{\cos \Delta x-1}{\Delta x}=0$
Since,
$\lim_{\Delta x\to 0}\frac{\cos \Delta x-1}{\Delta x}=0$
$\lim_{\Delta x\to 0}\sin x=\sin x$ (limit of constant function).
Therefore,
$\lim_{\Delta x\to 0}\sin x\left( \frac{\cos \Delta x -1}{\Delta x}\right)=0$
Because, "if two limits of two functions exist at a point then the limit of their product exists at the point and is equal to the product of their limits".
Similary, since,
$\lim_{\Delta x\to 0}\frac{\sin \Delta x}{\Delta x}=1$
$\lim_{\Delta x\to 0}\cos x=\cos x$ (limit of constant function).
Therefore,
$\lim_{\Delta x\to 0}\cos x\left( \frac{\sin \Delta x}{\Delta x} \right)=\cos x$
Because, "if two limits of two functions exist at a point then the limit of their products exists at the point and is equal to the product of their limits"
Thus,
$\lim_{\Delta x\to 0}\left[\sin x\left( \frac{\cos \Delta x-1}{\Delta x}\right)+\cos x\left( \frac{\sin \Delta x}{\Delta x}\right) \right]$= $0+\cos x=\cos x$
Because, "if the limits of two functions exist at a point then the limit of the sum of these functions exists at the point and is equal to the sum of the limits".

6. ## thank you

Oh... I 've got it!! Thanks a lot!