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anonymous
 5 years ago
(d/dx)[(xcosa)^2+2x(sinb)^2]=
anonymous
 5 years ago
(d/dx)[(xcosa)^2+2x(sinb)^2]=

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Dependin on whether you assume that a and b are functions of x, or constants, you get two very different answers. First we'll approach this with the assumption that both a and b are functions of x.  d/dx = 2(xcos(a)*([xsin(a)*da/dx+cos(a)]+[2sin(b)2xcos(b)*db/dx])/sin^2(b)   If you assume that both of the trig ratios are constants, then it ends up as d/dx= 2(xcos(a))*cos(a) +2/sin(b)
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