if im having to use the definition of the derivative $\displaystyle \frac{(x+h)-f(x)}{h} $ to show if f(x)=sin2x then f'(x)=2cos2x how would i do this.i can do it for normal non trig functions?thanks sum difference formula?
You'll need to use the comount angle forum on $\displaystyle \sin(2(x +h)) = \sin2x \cos 2h + \cos 2x \sin 2h$
and you'll need to use the following limits $\displaystyle \lim_{X \rightarrow 0} \left( \frac{\sin X}{X} \right) = 1$ and $\displaystyle \lim_{X \rightarrow 0} \left( \frac{1-\cos X}{X} \right) = 0$
Bobak
I made a little errror in my post above Its fixed now.
$\displaystyle f'(x) = \lim_{h \rightarrow 0} \left( \frac{\sin2x \cos 2h + \cos 2x \sin 2h-\sin 2x}{h} \right )
$
$\displaystyle \Rightarrow \lim_{h \rightarrow 0} \left( \frac{\sin2x (\cos 2h -1) + \cos 2x \sin 2h}{h} \right )$
multiply the top and bottom by 2
$\displaystyle \Rightarrow \lim_{h \rightarrow 0} \left( \frac{2\sin2x (\cos 2h -1) + 2\cos 2x \sin 2h}{2h} \right )$
Then continue.
Bobak
No don't apply any double angle formula.
take a minus out of the cosine limit part and split the sum into two limits.
$\displaystyle
\Rightarrow \lim_{h \rightarrow 0} \left( \frac{-2\sin2x (1 -\cos 2h)}{2h}\right ) + \lim_{h \rightarrow 0} \left( \frac{2\cos 2x \sin 2h}{2h} \right )
$
take anything that is independent of h outside the limit.
$\displaystyle
\Rightarrow -2\sin2x \lim_{h \rightarrow 0} \left( \frac{ (1 -\cos 2h)}{2h}\right ) + 2\cos 2x \lim_{h \rightarrow 0} \left( \frac{\sin 2h}{2h} \right )
$
Now just apply the two limits I gave you in my first post.
Bobak