A
y`=[(2x-3)`*√(x²+4x-3)-(√(x²+4x-3))`*(2x-3)]/(√(x²+4x-3))²=
=[2√(x²+4x-3)-(2x-3)(2x+4)/2√(x²+4x-3)]/(x²+4x-3)=
=[4(x²+4x-3)-(2x-3)(2x+4)]/2√(x²+4x-3)³=
=(4x²+16x-12-4x²+6x-8x+12)/2√(x²+4x-3)³=14x/2√(x²+4x-3)³=7x/√(x²+4x-3)³
б
y`=(3^arctg2x)`-[ln(1+4x²)]`=2ln3*3^arctg2x/(1+4x²)+8x/(1+4x²)
в
(lnx)`=1/x,поэтому когда делим на дробь,то ее переворачиваем
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