1. ## Basic Calculas (d/dx)

ok (d/dx) sin x= cos x

therefore (d/dx)cos x= -sinx

but how do u explain in terms how (d/dx)cosx = sin x

i believe i am supposed to us the formula f'(x)= F(x + h) - f(x) / h

but can someone show me how it is done so i can understand how it is expressed.

2. Ok. So the limit definition by your notation should be something like $f'(x)=\lim_{h \rightarrow 0} \frac{ \cos{(x+h)}-\cos{x}}{h}$. So to handle the numerator use this identity - $\cos{(a+b)}=\cos{(a)}\cos{(b)}-\sin{(a)}\sin{(b)}$. Apply this to your limit and post back what you find.

3. Thinking more that actually is a tricky limit. Try just using the fact that $
\sin{ \left( \frac{\pi}{2} + x \right)} = \cos{x}$
and that $\sin(a+b) = \sin(a) \sin(b) + \cos(a) \cos(b)$

4. $f'(x)=\lim_{h \rightarrow 0} \frac{ \cos{(x+h)}-\cos{x}}{h}$

$\lim_{h \rightarrow 0}\frac{cos(x)cos(h)-sin(x)sin(h) - cos(x)}{h}$

$\lim_{h \rightarrow 0}\frac{-sin(x)sin(h) -cos(x)(1- cos(h))}{h}$

$-\lim_{h \rightarrow 0}\frac{sin(x)sin(h)}{h} -\lim_{h \rightarrow 0}\frac{cos(x)(1-cos(h))}{h}$

$-sin(x)\lim_{h \rightarrow 0} \frac{sin(h)}{h}-cos(x)\lim_{h \rightarrow 0}\frac{1-cos(h)}{h}$

Then remember that

$\lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1$

and

$\lim_{h \rightarrow 0} \frac{1-cos(h)}{h}=0$

You should be able to finish up from here and that was a tricky question lol

5. ^ Well just do it all why don't you?

Nicely done. OP that's exactly what you want.

6. lol yep I was stuck for a good bit becuause the cos(x) in the numerator would cancel out leaving $\frac{0}{0}$

7. Yop !

A more straighforward method is to use $\cos(a)-\cos(b)=-2 \sin \frac{a+b}{2} \sin \frac{a-b}{2}$

$\frac{\cos(x+h)-\cos(x)}{h}=-2 \cdot \frac{\sin(x+\tfrac h2) \sin(\tfrac h2)}{h}=-2 \cdot \sin(x+\tfrac h2) \cdot \frac{\sin(\tfrac h2)}{h}$

We know that $\lim_{x \to 0} \frac{\sin(ax)}{x}=a.$ Therefore $\lim_{h \to 0} \frac{\sin(\tfrac h2)}{h}=\frac 12$

And we have $\lim_{h \to 0} \frac{\cos(x+h)-\cos(x)}{h}=-\sin(x)$