We first work out the derivative of sinx cosx
Using product rule:
ddxsinxcosx=(cosx)(cosx)+(−sinx)(sinx)=cos2x
Using chain rule:
y=√u,u=sinxcosx
dydx=dydu×dudx=12√u×cos2x=cos2x2√sinxcosx
Thanks Max, I am just going to do it a different way.
What's the derivative of √sinxcosx?
sin2x=2sinxcosxso0.5sin2x=sinxcosxddx[0.5sin2x]0.5=ddx√0.5[sin2x]0.5=√0.5∗0.5[sin2x]−0.5∗2cos(2x)=√0.5cos(2x)√sin(2x)=cos(2x)√2sin(2x)=cos(2x)√2∗2sinxcosc=cos(2x)2√sinxcosc
There you go - our answers are the same ... hopefully that means we are both correct !!!