Question

Question #128f6

Answer 1

`d/dx (cos ax) = lim_(h->0) (cos(a(x+h)) -cosax)/h`

`d/dx (cos ax) = lim_(h->0) (cos(ax+ah) -cosax)/h`

`d/dx (cos ax) = lim_(h->0) (cosax cos ah -sin ax sin ah + -cosax)/h`

`d/dx (cos ax) = lim_(h->0) (-cosax (1-cos ah) -sin ax sin ah + )/h`

`d/dx (cos ax) = -acos ax lim_(h->0) (1-cos ah)/(ah) -asin ax lim_(h->0) (sin ah )/(ah)`

Now:

`lim_(h->0) (sin ah )/(ah) =1`

`lim_(h->0) (1-cos ah)/(ah) = lim_(h->0) (2sin^2 ((ah)/2))/(ah) = lim_(h->0) sin ((ah)/2) xx sin((ah)/2)/((ah)/2) = 0*1=0`

so:

`d/dx (cos ax) = -acos ax * 0 -asin ax *1 = -asin ax`