Differentiating sine / cosine The sine / cosine functions differentiate using standard sets of rules:

tcos)t(sindtd

= tsin)t(cosdtd

−=

tcos)tsin(dtd

−=− tsin)tcos(dtd

=−

Slightly more complicated is sin (ωt) or sin (ωt + φ) etc. In these cases the expression within the sine or cosine is itself differentiated w.r.t the variable “t”, in addition to the sin or cos function itself: these are then multiplied together.

tcosA)tsinA(dtd ωωω = and tsinA)tcosA(

dtd ωωω −= since ωω =)t(

dtd

For a further example, if the sin / cos function is now a more complicated function of t say, notice how we still differentiate it (as well as the sin / cos itself) ; e.g. for sin (kt2) and cos (kt3) etc.

[ ] )ktcos(.kt)ktsin(dtd 22 2= and [ ] 323 3 ktsin.kt)ktcos(

dtd

−=

If the function has two variables in it, (as with some of the functions in our lecture course, e.g. ( )kxtsinyy −= ω0 ), then the differentiation works in the same way;

( )kxtcosy)y(dtd

−= ωω 0 and ( )kxtcosky)y(dxd

−−= ω0

In general: dxdg

dxdf [f(g(x))]

dxd

×=

This means we differentiate the outside function, leave the argument of the outside function alone, and then multiply by the derivative of the inside function. For example;

1)3xsin(.xx1)3xsin(dxdg

dxdf 1)][cos(3x

dxd 222 +−=×+−=×=+ 66

PHY1106: Waves and Oscillators: Dr. Pete Vukusic

Practice Problems: Complete the following (see PV or your tutor for solutions):

[ ] =)tsin(Adtd ωω

[ ] =− )kxtcos(Adxd ω

[ ] =− )kxtcos(Adtd ωω

=

−−

ωω )kxtcos(A

dtd

( )( ) =

22

/a/kasin.c

dkd