derivative of sinx (1 Viewer)

[fashionista]

hiyaa
we havent doen this(seriving trig functions) in class yet but i needed it for 4u questions.
i got a little bit confused around the 3rd or 4th line of deriving sinx where it went from
(lim h-0)(2cos(x+h/2)sin(h/2))/h
to
(lim h-0)(cos(x+h/2) * lim (h-0) (sin (h/2))/(h/2)
thats usinf first principles by the way
can sum1 pleese explain where the h and the 2 from the cos bit go?
wait....i think i get it.....is the whole thing just multiplied by the reciprocal of h/2? am i rite?

 

[Giant Lobster]

its just algebra
they took the 2 from the numerator to the denominator, and in doing so they reverse the sign of its power. i.e. 2 becomes 1/2 down under

 

[ND]

Originally posted by fashionista
(lim h-0)(cos(x+h/2) * lim (h-0) (sin (h/2))/(h/2)

For the lim h->0(cos(x+h/2), it's rpetty obvious that as h->0, cos(x+h/2)-->cos(x). So lim h->0(cos(x+h/2)=cosx.
For lim h->0 [sin(h/2)/(h/2)], remember that when an angle is very small, the angle approximately equals the sin of that angle. i.e. sin(h/2)=h/2, when h is very very small (as is the case as it approaches 0), and so lim h->0 [sin(h/2)/(h/2)]=1.

So:

(lim h-0)(cos(x+h/2) * lim (h-0) (sin (h/2))/(h/2)= cosx * 1
= cosx

edit: oops sorry i comletely misread your quesition.... yeh all the did was split it up, and divide by 1/2 (which is the same as multiplying by 2).

 

[fashionista]

thanks u!