
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.

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

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

$\displaystyle f'(x)=\lim_{h \rightarrow 0} \frac{ \cos{(x+h)}\cos{x}}{h}$
$\displaystyle \lim_{h \rightarrow 0}\frac{cos(x)cos(h)sin(x)sin(h)  cos(x)}{h}$
$\displaystyle \lim_{h \rightarrow 0}\frac{sin(x)sin(h) cos(x)(1 cos(h))}{h}$
$\displaystyle \lim_{h \rightarrow 0}\frac{sin(x)sin(h)}{h} \lim_{h \rightarrow 0}\frac{cos(x)(1cos(h))}{h}$
$\displaystyle sin(x)\lim_{h \rightarrow 0} \frac{sin(h)}{h}cos(x)\lim_{h \rightarrow 0}\frac{1cos(h)}{h}$
Then remember that
$\displaystyle \lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1$
and
$\displaystyle \lim_{h \rightarrow 0} \frac{1cos(h)}{h}=0$
You should be able to finish up from here and that was a tricky question lol

^ Well just do it all why don't you? (Rofl)
Nicely done. OP that's exactly what you want.

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

Yop !
A more straighforward method is to use $\displaystyle \cos(a)\cos(b)=2 \sin \frac{a+b}{2} \sin \frac{ab}{2}$
$\displaystyle \frac{\cos(x+h)\cos(x)}{h}=2 \cdot \frac{\sin(x+\tfrac h2) \sin(\tfrac h2)}{h}=2 \cdot \sin(x+\tfrac h2) \cdot \frac{\sin(\tfrac h2)}{h}$
We know that $\displaystyle \lim_{x \to 0} \frac{\sin(ax)}{x}=a.$ Therefore $\displaystyle \lim_{h \to 0} \frac{\sin(\tfrac h2)}{h}=\frac 12$
And we have $\displaystyle \lim_{h \to 0} \frac{\cos(x+h)\cos(x)}{h}=\sin(x)$