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d((cos((2*pi*t)/24)) -2y^2-4y+1.5)/dy any one know how to calculate this?

 May 8, 2016
 #1
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Possible derivation:
d/dy(1.5-4 y-2 y^2+cos((pi t)/12))
Differentiate the sum term by term and factor out constants:
  =  d/dy(1.5)-4 (d/dy(y))-2 (d/dy(y^2))+d/dy(cos((pi t)/12))
The derivative of 1.5 is zero:
  =  -4 (d/dy(y))-2 (d/dy(y^2))+d/dy(cos((pi t)/12))+0
Simplify the expression:
  =  -4 (d/dy(y))-2 (d/dy(y^2))+d/dy(cos((pi t)/12))
The derivative of y is 1:
  =  -2 (d/dy(y^2))+d/dy(cos((pi t)/12))-1 4
Use the power rule, d/dy(y^n) = n y^(n-1), where n = 2: d/dy(y^2) = 2 y:
  =  -4+d/dy(cos((pi t)/12))-2 2 y
Simplify the expression:
  =  -4-4 y+d/dy(cos((pi t)/12))
The derivative of cos((pi t)/12) is zero:
  =  -4-4 y+0
Simplify the expression:
Answer: |  =  -4-4y

 May 8, 2016

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