Find $\frac{d}{dx}(\cos x)$

Question

Find $\dfrac{d}{dx}(\cos x)$

I know the answer is $-\sin x$ only by process of elimination. I can find solution graphically but I need to know algebraically. Here is my proof so far.
$\begin{align*} \dfrac{d}{dx}\cos x=\lim_{h\to 0}\dfrac{\cos (x+h)-\cos x}{h} &=\lim_{h\to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h} \end{align*}$

And that's where I end up and I have no clue where to go from here. Can someone please give me the next step but not the complete answer.

Answer 1

You need to know how to evaluate the following limits: $$ \lim_{h \rightarrow 0} \frac{ \cos{h} - 1}{h}, \quad \lim_{h \rightarrow 0} \frac{ \sin{h}}{h}$$

Answer 2

sorry i cant and i dont know how to write math symbols here hope u understand like that.

lim [ cos(x + h) - cos(x) ] / h h -> 0

lim [ cos(x)cos(h) - sin(x)sin(h) - cos(x) ] / h h -> 0

lim [ cos(x)cos(h) - cos(x) - sin(x)sin(h) ] / h h -> 0

lim [ cos(x) [ cos(h) - 1 ] - sin(x)sin(h) ] / h h -> 0

lim [ cos(x) [ cos(h) - 1 ]/h - [sin(x)sin(h)]/h h -> 0

lim [ cos(x) [cos(h) - 1]/h ] - lim sin(x)sin(h)/h h -> 0 . . . . . . . . . . . . . . . . . . h -> 0

Factor cos(x) from the first limit, and sin(x) from the second limit.

cos(x) lim [ cos(h) - 1 ]/h - sin(x) lim sin(h)/h . . . . . . h -> 0 . . . . . . . . . . . . . . . . . h -> 0

We need foreknowledge of trig limits to realize that lim [ cos(h) - 1 ]/h = 0, and h -> 0 lim sin(h)/h = 1 h -> 0

The above then becomes

cos(x) (0) - sin(x)(1) 0 - sin(x)

     or

cos x*cos h - sin x*sin h -cos x)/h As h is tending to zero..... cos h =1 =(cos x - cos x -sin x*sin h)/h = -sin x*(sin h)/h sin h/h=1 = - sin x

-sin(x)

Answer 3

If you know $d \sin x / dx$, then relate $\cos$ to $\sin$.

Or find some other way to combine things to relate what you know to what you don't know (or what you don't know to itself). One mildly amusing approach is

$$ \frac{d}{dx} \left( \sin^2 x + \cos^2 x \right) $$

Answer 4

$$\cos h -1=\frac{(\cos h -1)(\cos h+1)}{\cos h +1}$$

Alternately, use the fact that $$\cos 2t=\cos^2 t-\sin^2 t=2\cos^2 t-1=1-2\sin^2 t.$$