How can i prove cos2x derivation using cos x by basic derivation proof.... like (limit and delta(x) delta(y))
sheesh
$\displaystyle \frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}=$
$\displaystyle \frac{\cos(2x+2h)-cos(2x)}{h}=-2\sin(2x)$
$\displaystyle \frac{\cos(2x)\cos(2h)-\sin(2x)\sin(2h)-\cos(2x)}{h}=$
$\displaystyle \frac{\cos(2x)(cos(2h)-1)-\sin(2x)\sin(2h)}{h}=$
$\displaystyle \cos(2x)\frac{cos(2h)-1}{h}-\sin(2x)\frac{\sin(2h)}{h}$
we need to reduce the sin and cos terms that involve h further so we can use known limits
$\displaystyle \cos(2h)=2\cos^2(h)-1$ and $\sin(2h)=2\sin(h)\cos(h)$
so
$\displaystyle \displaystyle{\lim_{h\to 0}}\frac{\cos(2h)-1}{h}=\displaystyle{\lim_{h\to 0}}\frac{2\cos^2(h)-2}{h}=0$
$\displaystyle \displaystyle{\lim_{h\to 0}}\frac{\sin(2h)}{h}=\displaystyle{\lim_{h\to 0}}\frac{2\sin(h)\cos(h)}{h}=2$
and thus
$\displaystyle \displaystyle{\lim_{h\to 0}}\frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}=-2\sin(2x)$
It depends on what you already know. Are you allowed to use the Chain Rule? If so, you can prove $\displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left[ \cos{(x)} \right] = -\sin{(x)} \end{align*}$ by first principles and then apply the chain rule. If you really wanted to, you could prove $\displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left[ \sin{(x)} \right] = \cos{(x)} \end{align*}$ by first principles, then remember $\displaystyle \displaystyle \begin{align*} \cos{(x)} \equiv \sin{ \left( \frac{\pi}{2} - x \right) } \end{align*}$ and use the chain rule from there...