How do you differentiate # sin^2(x/2)#?

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Answer 1 · 2016-09-18T14:53:08

Letting #y = sin^2(u)# and #u = x/2#, we need to differentiate both functions and multiply the derivatives together.

The derivative of #y = sin^2u# can be obtained as follows:

#y = (sinu)(sinu)#

#y' = cosu xx sinu + cosu xx sinu#

#y' = 2cosusinu#

#y' = sin2u#

The derivative of #u = x/2# can be obtained using the quotient rule:

#u = x/2#

#u' = (1 xx 2 - x xx 0)/2^2#

#u' = 2/4#

#u' = 1/2#

So, the derivative of #y = sin^2(x/2)# is:

#dy/dx = sin2u xx 1/2#

#dy/dx = sin2(x/2) xx 1/2#

#dy/dx = 1/2 xx 2 xx sin(x/2) xx cos(x/2)#

#dy/dx = sin(x/2) xx cos(x/2)#

Hopefully this helps!

Answer 2 · 2016-09-18T15:02:26

#1/2sinx#.

We use the Trigo. Identity # : 1-costheta=2sin^2(theta/2)#.

So, in our case, #sin^2(x/2)=1/2(1-cosx)#. Hence,

#d/dx[sin^2(x/2)]=d/dx[1/2(1-cosx)]=1/2d/dx[1-cosx]#

#=1/2[d/dx1-d/dxcosx]=1/2[0-(-sinx)]=1/2sinx#.

Note that, since, #sintheta=2sin(theta/2)cos(theta/2)#

#d/dx[sin^2(x/2)=1/2sinx=1/2(2sin(x/2)cos(x/2))=sin(x/2)cos(x/2),#

just to match With theAnswer furnished by, HSBC244 !

Enjoy Maths!