Expert: NUR ARINA BASILAH BINTI KAMISAN - 8/26/2008

Question
please help me to find out the derivatives of the following problems: find dy/dt a)y=sqrt1+cost^2, b)y=(1+tan^4(t/2)^3,  c)y= sin^2((pi)t-2), d) y=(1+cos2t)^-4 thank you

Answer
dear Ahmed,

when you derive trigonometry function,just derive it like a normal function.

a) y=[1+cost^2]^1/2
so dy/dt=1/2(1+cost^2)^-1/2*(-sint^2)*2t
simplify the equation above and you will have:
dy/dt=(-tsint^2)/[(1+cost^2)^1/2]

b)y=1+tan^4(t/2)^3
dy/dt=4tan^3(t/2)^3*sec^2(t/2)^3*3(t/2)^2
simplify the equation and you will have:
dy/dt=12(t/2)^2*sec^2(t/2)*tan^3(t/2)

c)y=sin^2((pi)t-2)
dy/dt=2sin((pi)t-2)*cos((pi)t-2)*pi
simplify the equation above and you will have:
dy/dt=2(pi)sin((pi)t-2)cos((pi)t-2)

d)y=(1+cos2t)^-4
dy/dt=-4(1+cos2t)^-5*(-2sin2t)
dy/dt=8sin2t(1+cos2t)^-5

good luck and try to write it back in mathematical form and understand it..
you can also look at this website for further information.
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/trigderivsoldirectory/TrigDe...