Đáp án:

f) \(\dfrac{{ - 4{x^2} - 12}}{{{{\left( {{x^2} - 2x - 3} \right)}^2}}}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)y' = \dfrac{{ - 2.3}}{{{{\left( {2x + 1} \right)}^2}}} =  - \dfrac{6}{{{{\left( {2x + 1} \right)}^2}}}\\
b)y' = \dfrac{{2\left( {1 - 3x} \right) + 3\left( {2x + 1} \right)}}{{{{\left( {1 - 3x} \right)}^2}}}\\
 = \dfrac{{2 - 6x + 6x + 3}}{{{{\left( {1 - 3x} \right)}^2}}} = \dfrac{5}{{{{\left( {1 - 3x} \right)}^2}}}\\
c)y' = \dfrac{{\left( { - 1 - 2x} \right)\left( {1 - x + {x^2}} \right) - \left( { - 1 + 2x} \right)\left( {1 + x - {x^2}} \right)}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}\\
 = \dfrac{{ - 1 + x - {x^2} - 2x + 2{x^2} - 2{x^3} - \left( { - 1 + 2x - x + 2{x^2} + {x^2} - 2{x^3}} \right)}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}\\
 = \dfrac{{ - 2x - 2{x^2}}}{{{{\left( {1 - x + {x^2}} \right)}^2}}}\\
e)y' = \dfrac{{\left( {4x - 4} \right)\left( {x - 3} \right) - 2{x^2} + 4x - 1}}{{{{\left( {x - 3} \right)}^2}}}\\
 = \dfrac{{4{x^2} - 16x + 12 - 2{x^2} + 4x - 1}}{{{{\left( {x - 3} \right)}^2}}}\\
 = \dfrac{{2{x^2} - 12x + 11}}{{{{\left( {x - 3} \right)}^2}}}\\
d)y' = \dfrac{{\left( {2x - 3} \right)\left( {x - 1} \right) - {x^2} + 3x - 3}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{2{x^2} - 5x + 3 - {x^2} + 3x - 3}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{{x^2} - 2x}}{{{{\left( {x - 1} \right)}^2}}}\\
f)y' = \dfrac{{4x\left( {{x^2} - 2x - 3} \right) - \left( {2x - 2} \right).2{x^2}}}{{{{\left( {{x^2} - 2x - 3} \right)}^2}}}\\
 = \dfrac{{4{x^3} - 8{x^2} - 12 - 4{x^3} + 4{x^2}}}{{{{\left( {{x^2} - 2x - 3} \right)}^2}}}\\
 = \dfrac{{ - 4{x^2} - 12}}{{{{\left( {{x^2} - 2x - 3} \right)}^2}}}
\end{array}\)

a, $y'=-\dfrac{3(2x+1)'}{(2x+1)^2}=\dfrac{-6x}{(2x+1)^2}$

b, $y'=\dfrac{(2x+1)'(1-3x)-(2x+1)(1-3x)'}{(1-3x)^2}=\dfrac{2(1-3x)+3(2x+1)}{(1-3x)^2}=\dfrac{5}{(3x-1)^2}$

c, $y'=\dfrac{(1-x-x^2)'(1-x+x^2)-(1-x-x^2)(1-x+x^2)'}{(1-x+x^2)^2}=\dfrac{(x^2-x+1)(-2x-1)+(x^2+x-1)(2x+1)}{(x^2-x+1)^2}$

d, $y'=\dfrac{(x^2-3x+3)'(x-1)-(x^2-3x+3)(x-1)'}{(x-1)^2}=\dfrac{(2x-3)(x-1)-x^2+3x-3}{(x-1)^2}=\dfrac{2x^2-5x+3-x^2+3x-3}{(x-1)^2}=\dfrac{x^2-2x}{(x-1)^2}$

e, $y'=\dfrac{(2x^2-4x+1)'(x-3)-(2x^2-4x+1)(x-3)'}{(x-3)^2}=\dfrac{(4x-4)(x-3)-2x^2+4x-1}{(x-3)^2}=\dfrac{4x^2-16x+12-2x^2+4x-1}{(x-3)^2}=\dfrac{2x^2-12x+11}{(x-3)^2}$

f, $y'=\dfrac{(2x^2)'(x^2-2x-3)-2x^2(x^2-2x-3)'}{(x^2-2x-3)^2}=\dfrac{4x(x^2-2x-3)-2x^2(2x-2)}{(x^2-2x-3)^2}=\dfrac{-4x^2-12x}{(x^2-2x-3)^2}$