Proove, from the definition of a differential, that if f(x) = sinax, f'(x) = acosax?

Question

Im soo close to figuring it out, i got my notepad filled with math but im still not getting it. If you don't know then give me some useful links which can help me. THANKS!

Answer 1

I assume you mean "definition of the derivative"?

f'(x) = lim h->0 [sin(a(x+h)) - sin(ax)] / h
= lim h->0 [sin(ax)cos(ah)+cos(ax)sin(ah)-sin(ax)]/h
= lim h->0 sin(ax)[cos(ah)-1]/h + cos(ax)sin(ah)/h.

Now,

lim h->0 sin(ah)/h =
lim h->0 (a/a)sin(ah)/h =
lim h->0 a*[sin(ah)/(ah)] =
a*lim h->0 [sin(ah)/(ah)] =
a*1 = a, because lim x->0 sin(x)/x=1.

Also,

lim h->0 [cos(ah)-1]/h=
lim h->0 (cos(ah)-1)(cos(ah)+1)/[h(cos(ah)+1)] =
lim h->0 (cos^2(ah)-1)/[h(cos(ah)+1)] =
lim h->0 -sin^2(ah)/[h(cos(ah)+1)] =
lim h->0 -[sin(ah)/h][sin(ah)/(cos(ah)+1)] =
lim h->0 -[a*sin(ah)/(ah)][sin(ah)/(cos(ah)+1)] =
-lim h->0 a*sin(ah)/(cos(ah)+1) =
-(a/2) lim h->0 sin(ah) = 0
beacuse sin(0)=0.

Put it all together, and you get f'(x)=a*cos(ax).

Answer 2

f(x)=sinax
diff wrt x we get
f'(x)=cosax.a(here a is constant so on differentiating wrt x we get again a)
therefore we get f'(x)=a cos ax