1. ## differentiation quiery

if im having to use the definition of the derivative $\frac{(x+h)-f(x)}{h}$ 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?

2. Originally Posted by skystar
if im having to use the definition of the derivative $\frac{(x+h)-f(x)}{h}$ 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

You'll need to use the comount angle forum on $\sin(2(x +h)) = \sin2x \cos 2h + \cos 2x \sin 2h$

and you'll need to use the following limits $\lim_{X \rightarrow 0} \left( \frac{\sin X}{X} \right) = 1$ and $\lim_{X \rightarrow 0} \left( \frac{1-\cos X}{X} \right) = 0$

Bobak

3. ive expanded it outand put over h, i get cos2x im a bit confused about the cos(h)/h part i think. thanks

4. $\frac{sin2x(cos2h-1)+cos2xsin2h}{h}$ thats where im at and now im just stuck,thanks,

5. Originally Posted by skystar
ive expanded it outand put over h, i get cos2x im a bit confused about the cos(h)/h part i think. thanks
I made a little errror in my post above Its fixed now.

$f'(x) = \lim_{h \rightarrow 0} \left( \frac{\sin2x \cos 2h + \cos 2x \sin 2h-\sin 2x}{h} \right )
$

$\Rightarrow \lim_{h \rightarrow 0} \left( \frac{\sin2x (\cos 2h -1) + \cos 2x \sin 2h}{h} \right )$

multiply the top and bottom by 2

$\Rightarrow \lim_{h \rightarrow 0} \left( \frac{2\sin2x (\cos 2h -1) + 2\cos 2x \sin 2h}{2h} \right )$

Then continue.

Bobak

6. after
$
\Rightarrow \lim_{h \rightarrow 0} \left( \frac{2\sin2x (\cos 2h -1) + 2\cos 2x \sin 2h}{2h} \right )
$

im still confused i can make 2sin2x into things like 4sinxcosx but i cant see a way to simplify.

7. Originally Posted by skystar
after
$
\Rightarrow \lim_{h \rightarrow 0} \left( \frac{2\sin2x (\cos 2h -1) + 2\cos 2x \sin 2h}{2h} \right )
$

im still confused i can make 2sin2x into things like 4sinxcosx but i cant see a way to simplify.
No don't apply any double angle formula.

take a minus out of the cosine limit part and split the sum into two limits.

$
\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.

$
\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