Derivative of sine function

[Christina]

hi
plz show me some of steps on differentiating g(x)=2sinx, because i understand that the derivative is g'(x)=2cosx....but i don't really understand why. Is there a simplier way to differentiate it without using limits and First principles?

thank u

 

[stapel]

Is there a simplier way to differentiate it without using limits and First principles?

Not if you're still at that point in the book where you're supposed to be using limits and such. Where are you stuck in the limit process?

Please be specific. Thank you.

Eliz.

 

[skeeter]

to understand "why" the derivative of 2sinx is 2cosx, you need to use the definition of a derivative, and ,yes, that involves limits.

f(x) = 2sin(x)

f(x+h) = 2sin(x+h) = 2(sinx*cosh + cosx*sinh)

lim{h->0} [f(x+h) - f(x)]/h =

lim{h->0} (2sinx*cosh + 2cosx*sinh - 2sinx)/h =

lim{h->0} (2sinx*cosh - 2sinx + 2cosx*sinh)/h =

2*lim{h->0} (sinx*cosh - sinx + cosx*sinh)/h =

2*lim{h->0} [sinx(cosh - 1) + cosx*sinh)/h =

2*lim{h->0} sinx(cosh - 1)/h + cosx*sinh/h =

2*lim{h->0} sinx(cosh - 1)/h + 2*lim{h->0} cosx*sinh/h =

2*sinx*(0) + 2*cosx*(1) = 2cosx